Free math worksheets

- Identify \(A\), \(B\), and \(C\) from the quadratic terms. - The rotation formula gives the cotangent of twice the desired angle. - Choose the value of \(2\theta\) that places \(\theta\) in the stated interval.

1. The coefficients are \(A=7\), \(B=6\sqrt{3}\), and \(C=1\). 2. Thus \(\cot(2\theta)=\frac{7-1}{6\sqrt{3}}=\frac{1}{\sqrt{3}}\). 3. In the required range, \(2\theta=60^\circ\), so \(\theta=30^\circ\).

\(\theta=30^\circ\)

- Substitute the original coordinates directly into the given rotation formulas. - Use the exact sine and cosine values for \(30^\circ\). - The opposite endpoint is the negative of the first because the ellipse is centered at the origin.

1. Substitute \(P=(3\sqrt{3}, 3)\): \(x'=(3\sqrt{3})(\frac{\sqrt{3}}{2})+3(\frac{1}{2})=6\). 2. Its second coordinate is \(y'=-(3\sqrt{3})(\frac{1}{2})+3(\frac{\sqrt{3}}{2})=0\). 3. Thus \(P\) has rotated coordinates \((6, 0)\). 4. The opposite endpoint is \((-3\sqrt{3}, -3)\), which transforms to \((-6, 0)\). 5. Both endpoints lie on the \(x'\)-axis, confirming that the rotation aligns that axis with the ellipse’s major axis.

The major-axis endpoints have rotated coordinates \((6, 0)\) and \((-6, 0)\).

- Separate a geometric object from the coordinate numbers used to describe it. - Ask what is physically drawn in the plane and what depends on the chosen reference directions. - Compare the solid and dashed axis directions without imagining that the ellipse itself has been redrawn.

1. Changing the coordinate axes changes the reference directions used to assign coordinates; it does not apply a physical transformation to the objects in the drawing. 2. Therefore the ellipse's geometric shape and size stay the same. 3. The physical location of \(P\) also stays the same. 4. The numerical coordinate pair assigned to \(P\) may change because its components are measured along different axis directions.

a) The ellipse's geometric shape and size stay the same. b) The physical location of \(P\) stays the same. c) The coordinate pair assigned to \(P\) may change. Changing coordinate axes changes the description of fixed geometric objects, not the objects themselves.

- Compare the dashed principal axes with the horizontal and vertical coordinate axes. - Recall what the mixed \(xy\)-term signals about the alignment of a noncircular conic's principal axes.

1. The dashed principal axes are diagonal rather than parallel to the \(x\)- and \(y\)-axes, so the ellipse is rotated relative to the displayed coordinate system. 2. For a noncircular ellipse whose principal axes are rotated relative to the coordinate axes, expanding its principal-form equation in \(x\) and \(y\) generally produces a nonzero \(xy\)-term.

The ellipse is rotated relative to the \(x\)- and \(y\)-axes, and a nonzero \(xy\)-term is generally expected.

- When the coefficients of \(x^2\) and \(y^2\) are equal, the rotation-angle formula simplifies immediately. - Substitute the \(45^\circ\) coordinate formulas and combine the squared-coordinate terms. - Normalize the rotated equation so the right side is \(1\) before classifying.

1. Here \(A=C=8\), so \(\cot(2\theta)=\frac{A-C}{B}=0\). Choose \(2\theta=90^\circ\), giving \(\theta=45^\circ\). 2. Use \(x=\frac{x'-y'}{\sqrt{2}}\) and \(y=\frac{x'+y'}{\sqrt{2}}\). 3. Then \(x^2+y^2=x'^2+y'^2\) and \(xy=\frac{x'^2-y'^2}{2}\). 4. The equation becomes \(11x'^2+5y'^2=25\). 5. In standard form, \(\frac{x'^2}{25/11}+\frac{y'^2}{5}=1\). Both squared terms are positive, so the conic is an ellipse.

A valid rotation is \(\theta=45^\circ\). Principal form: \(\frac{x'^2}{25/11}+\frac{y'^2}{5}=1\) Classification: ellipse

- Rewrite the expression using \(x^2+y^2\) and \(xy\) under the given rotation. - Confirm that no mixed principal-coordinate term remains before normalizing. - The larger denominator in principal form identifies the major-axis direction.

1. Under the given rotation, \(x^2+y^2=x'^2+y'^2\) and \(xy=\frac{x'^2-y'^2}{2}\). 2. Substitute into the quadratic: \(9(x'^2+y'^2)-4(x'^2-y'^2)=10\). 3. Combine like terms to obtain \(5x'^2+13y'^2=10\). 4. The principal standard form is \(\frac{x'^2}{2}+\frac{y'^2}{10/13}=1\). 5. The larger denominator is under \(x'^2\), so the major axis is the \(x'\)-axis. In the original plane, that direction is \(y=x\).

Principal standard form: \(\frac{x'^2}{2}+\frac{y'^2}{10/13}=1\) Major-axis direction: \(y=x\)

- Substitute both rotation formulas before simplifying. - Watch what happens to the two mixed-product terms. - Interpret the final equation geometrically for arbitrary \(\theta\).

1. Substitute the rotation formulas into \(x^2+y^2\). 2. Expanding gives \(x'^2\cos^2\theta-2x'y'\sin\theta\cos\theta+y'^2\sin^2\theta+x'^2\sin^2\theta+2x'y'\sin\theta\cos\theta+y'^2\cos^2\theta\). 3. The mixed terms cancel, and \(\sin^2\theta+\cos^2\theta=1\), so \(x^2+y^2=x'^2+y'^2\). 4. Therefore, the rotated equation is \(x'^2+y'^2=R^2\), identical in form to the original equation. 5. Because every rotation produces the same equation, every pair of perpendicular diameters can serve as principal axes; no single direction is preferred.

The rotated equation is \(x'^2+y'^2=R^2\). A centered circle has no unique principal-axis direction because its equation and geometry are unchanged by every rotation about its center.

- Read \(A\), \(B\), and \(C\) from the quadratic terms. - Convert the cotangent equation to a tangent equation for \(2\theta\). - Divide the resulting acute angle by \(2\).

1. Here \(A=12\), \(B=10\), and \(C=3\). 2. Thus \(\cot(2\theta)=\frac{12-3}{10}=\frac{9}{10}\), so \(\tan(2\theta)=\frac{10}{9}\). 3. The smallest positive solution is \(2\theta=\arctan(\frac{10}{9})\). 4. Therefore \(\theta=\frac{1}{2}\arctan(\frac{10}{9})\approx24.0^\circ\).

\(\theta=\frac{1}{2}\arctan(\frac{10}{9})\approx24.0^\circ\)

- Identify \(A\), \(B\), and \(C\), then apply the course formula for \(\cot(2\theta)\). - When \(A=C\) and \(B\ne0\), a \(45^\circ\) axis rotation is a natural candidate. - Substitute the rotated-coordinate formulas and normalize the equation after the mixed term cancels.

1. Here \(A=5\), \(B=-6\), and \(C=5\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}\): \(\cot(2\theta)=0\). 3. Choose \(2\theta=\frac{\pi}{2}\), so \(\theta=\frac{\pi}{4}\). 4. Substitute \(x=u\cos\theta-v\sin\theta\) and \(y=u\sin\theta+v\cos\theta\). 5. The mixed terms cancel, and the equation becomes \(2u^2+8v^2=20\). 6. Dividing by \(20\) gives \(\frac{u^2}{10}+\frac{v^2}{5/2}=1\), so the conic is an ellipse.

Rotation angle: \(\theta=\frac{\pi}{4}\) Rotated equation: \(\frac{u^2}{10}+\frac{v^2}{5/2}=1\) Conic: ellipse

- Project the original coordinate vector onto the rotated principal axes. - Substitute both principal-coordinate expressions into the standard ellipse equation. - Combine like terms carefully, especially the two contributions to the mixed term.

1. The principal coordinates are \(x'=x\cos30^\circ+y\sin30^\circ\) and \(y'=-x\sin30^\circ+y\cos30^\circ\). 2. Thus, \(x'=\frac{\sqrt{3}}{2}x+\frac{1}{2}y\) and \(y'=-\frac{1}{2}x+\frac{\sqrt{3}}{2}y\). 3. Substitute these expressions into \(\frac{x'^2}{25}+\frac{y'^2}{9}=1\). 4. Expanding and multiplying by \(225\) gives \(13x^2-8\sqrt{3}xy+21y^2=225\).

\(13x^2-8\sqrt{3}xy+21y^2=225\)

- Express the principal coordinates as projections onto axes rotated by \(45^\circ\). - Substitute those coordinates into the principal hyperbola equation before expanding. - Clear denominators first to reduce arithmetic errors in the mixed term.

1. The principal coordinates are \(x'=\frac{x+y}{\sqrt{2}}\) and \(y'=\frac{-x+y}{\sqrt{2}}\). 2. Substitute into the principal equation: \(\frac{(x+y)^2}{18}-\frac{(-x+y)^2}{8}=1\). 3. Multiply by \(72\): \(4(x+y)^2-9(-x+y)^2=72\). 4. Expanding and combining like terms gives \(-5x^2+26xy-5y^2=72\).

\(-5x^2+26xy-5y^2=72\)

- Use the course rotation formula to determine the double angle. - Substitute the standard axis-rotation formulas and verify cancellation of the mixed term. - Match the larger denominator with the corresponding rotated axis direction.

1. Here \(A=7\), \(B=4\sqrt{3}\), and \(C=3\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}=\frac{1}{\sqrt{3}}\). 3. Choose \(2\theta=\frac{\pi}{3}\), so \(\theta=\frac{\pi}{6}\). 4. Substitute \(x=x'\cos\theta-y'\sin\theta\) and \(y=x'\sin\theta+y'\cos\theta\). 5. The equation becomes \(9x'^2+y'^2=36\), or \(\frac{x'^2}{4}+\frac{y'^2}{36}=1\). 6. The \(x'\)-axis is \(30^\circ\) counterclockwise from the x-axis, and the \(y'\)-axis is perpendicular to it. The major axis is the \(y'\)-axis.

Rotation angle: \(\theta=\frac{\pi}{6}\) Principal form: \(\frac{x'^2}{4}+\frac{y'^2}{36}=1\) Minor-axis direction: \(30^\circ\); major-axis direction: \(120^\circ\) from the positive x-axis

- Equal x-squared and y-squared coefficients make a \(45^\circ\) rotation a natural choice. - Substitute the standard \(45^\circ\) coordinate formulas and combine the quadratic terms. - Normalize the result so the right side is \(1\) before classifying.

1. Here \(A=5\), \(B=8\), and \(C=5\). 2. Since \(A=C\) and \(B\ne0\), choose \(\theta=\frac{\pi}{4}\). 3. Substitute \(x=\frac{u-v}{\sqrt{2}}\) and \(y=\frac{u+v}{\sqrt{2}}\). 4. The mixed term cancels, and the equation becomes \(9u^2+v^2=18\). 5. Therefore, the principal form is \(\frac{u^2}{2}+\frac{v^2}{18}=1\), which is an ellipse.

Rotation angle: \(\theta=\frac{\pi}{4}\) Principal form: \(\frac{u^2}{2}+\frac{v^2}{18}=1\) Conic: ellipse

- Use the standard \(45^\circ\) coordinate formulas for \(x\) and \(y\). - The product of a sum and difference becomes a difference of squares. - Match the rotated u- and v-axes with their directions in the original coordinate plane.

1. For a \(45^\circ\) counterclockwise axis rotation, use \(x=\frac{u-v}{\sqrt{2}}\) and \(y=\frac{u+v}{\sqrt{2}}\). 2. Then \(xy=\frac{(u-v)(u+v)}{2}=\frac{u^2-v^2}{2}\). 3. The equation \(xy=4\) becomes \(u^2-v^2=8\). 4. In standard form, \(\frac{u^2}{8}-\frac{v^2}{8}=1\). 5. Thus, the u-axis and v-axis are the principal axes; in the original plane they lie along \(y=x\) and \(y=-x\).

Rotated equation: \(\frac{u^2}{8}-\frac{v^2}{8}=1\) Principal-axis directions in the original plane: \(y=x\) and \(y=-x\)

- The formulas given in the problem express old coordinates in terms of new coordinates. - To recover the new coordinates, invert the rotation rather than applying the same signs again. - A rotation matrix is inverted by reversing the rotation angle.

1. The given forward rotation relation is \(\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}x''\\y''\end{pmatrix}\). 2. Its inverse uses the transpose rotation matrix, so \(x''=x\cos\theta+y\sin\theta\) and \(y''=-x\sin\theta+y\cos\theta\). 3. With \(x=\sqrt{3}\), \(y=1\), and \(\theta=30^\circ\), \(x''=\sqrt{3}\cdot\frac{\sqrt{3}}{2}+1\cdot\frac{1}{2}=2\). 4. Also, \(y''=-\sqrt{3}\cdot\frac{1}{2}+1\cdot\frac{\sqrt{3}}{2}=0\). 5. Levent reused the forward rotation signs instead of inverting the coordinate transformation.

\((x'',y'')=(2,0)\) Levent used the forward rotation signs for the inverse transformation. The correct inverse formulas are \(x''=x\cos\theta+y\sin\theta\) and \(y''=-x\sin\theta+y\cos\theta\).

- Test the equation on each diagonal line separately. - Compare the distance from the origin to the ellipse along the two candidate principal axes. - The direction with the larger center-to-ellipse distance is the major-axis direction.

1. Along \(y=x\), substitute \(y=x\): \(5x^2-6x^2+5x^2=32\), so \(4x^2=32\) and \(x=\pm2\sqrt{2}\). The distance from the origin is \(\sqrt{x^2+y^2}=4\). 2. Along \(y=-x\), substitute \(y=-x\): \(5x^2+6x^2+5x^2=32\), so \(16x^2=32\) and \(x=\pm\sqrt{2}\). The distance from the origin is \(2\). 3. Therefore the major axis lies along \(y=x\), so diagram a) matches.

Diagram a); the ellipse extends farther along \(y=x\) than along \(y=-x\).

- Substitute the given coordinate relationships into every quadratic term before combining like terms. - After the mixed term disappears, normalize the equation so the right side is \(1\). - Convert only the two major-axis endpoints back to the original coordinates.

1. Substitute the rotation formulas into the quadratic expression to obtain \(5x^2-6xy+5y^2=2x'^2+8y'^2\). 2. The rotated equation is \(2x'^2+8y'^2=8\), or \(\frac{x'^2}{4}+y'^2=1\). 3. The major axis lies on the \(x'\)-axis and has endpoints \((x',y')=(2,0)\) and \((-2,0)\). 4. Converting these points back gives \((\sqrt{2},\sqrt{2})\) and \((-\sqrt{2},-\sqrt{2})\).

In rotated coordinates, the ellipse is \(\frac{x'^2}{4}+y'^2=1\). The major-axis endpoints in the original coordinates are \((\sqrt{2},\sqrt{2})\) and \((-\sqrt{2},-\sqrt{2})\).

- Read \(A\), \(B\), and \(C\) from the general quadratic before using the rotation-angle formula. - Choose the angle in the stated interval that has the required value of \(\cot(2\theta)\). - Substitute the rotated-coordinate formulas and confirm that the mixed term cancels before normalizing the equation.

1. Here \(A=7\), \(B=6\sqrt{3}\), and \(C=13\), so \(\cot(2\theta)=\frac{7-13}{6\sqrt{3}}=-\frac{1}{\sqrt{3}}\). 2. In the required range, \(2\theta=120^\circ\), so \(\theta=60^\circ\). 3. Substitute \(x=\frac{1}{2}x'-\frac{\sqrt{3}}{2}y'\) and \(y=\frac{\sqrt{3}}{2}x'+\frac{1}{2}y'\). After combining like terms, the equation becomes \(16x'^2+4y'^2=16\). 4. Divide by \(16\) to obtain \(x'^2+\frac{y'^2}{4}=1\). 5. Both squared terms have positive coefficients, so the conic is an ellipse.

Rotation angle: \(\theta=60^\circ\) Principal form: \(x'^2+\frac{y'^2}{4}=1\) Classification: ellipse

- Express the principal coordinates as projections onto the rotated axes. - Substitute those coordinate expressions before expanding. - Use the larger denominator in principal form to identify the major-axis direction.

1. For a \(30^\circ\) rotation, \(x'=x\cos30^\circ+y\sin30^\circ=\frac{\sqrt{3}x+y}{2}\) and \(y'=-x\sin30^\circ+y\cos30^\circ=\frac{-x+\sqrt{3}y}{2}\). 2. The principal equation is equivalent to \(4x'^2+9y'^2=36\). 3. Substitute the expressions for \(x'\) and \(y'\). 4. Expanding and simplifying gives \(21x^2-10\sqrt{3}xy+31y^2=144\). 5. Since the larger semi-axis is along the \(x'\)-axis, the major axis makes a \(30^\circ\) angle with the positive x-axis.

\(21x^2-10\sqrt{3}xy+31y^2=144\) The major axis is directed \(30^\circ\) counterclockwise from the positive x-axis.

- Use the mixed-term coefficient to determine twice the rotation angle. - Substitute the corresponding rotated-coordinate formulas before simplifying. - The larger denominator in principal form identifies the major-axis direction.

1. With \(A=3\), \(B=2\sqrt{3}\), and \(C=5\), the rotation angle satisfies \(\tan(2\theta)=\frac{B}{A-C}=-\sqrt{3}\). 2. Choose \(2\theta=-60^\circ\), so \(\theta=-30^\circ\). 3. Use \(x=\frac{\sqrt{3}}{2}x'+\frac{1}{2}y'\) and \(y=-\frac{1}{2}x'+\frac{\sqrt{3}}{2}y'\). 4. Substitution simplifies the equation to \(2x'^2+6y'^2=8\), or \(\frac{x'^2}{4}+\frac{y'^2}{4/3}=1\). 5. The larger denominator is under \(x'^2\), so the major axis follows the \(x'\)-axis, directed \(30^\circ\) clockwise from the positive x-axis.

One valid rotation is \(\theta=-30^\circ\). Principal form: \(\frac{x'^2}{4}+\frac{y'^2}{4/3}=1\) Major-axis direction: \(30^\circ\) clockwise from the positive x-axis

- Substitute the coordinate formulas for a \(45^\circ\) rotation into the product \(xy\). - Read the principal-axis features from the resulting standard hyperbola. - Convert each principal-coordinate point back to the original plane.

1. For a \(45^\circ\) rotation, use \(x=\frac{x'-y'}{\sqrt{2}}\) and \(y=\frac{x'+y'}{\sqrt{2}}\). 2. Then \(xy=\frac{x'^2-y'^2}{2}\), so the equation becomes \(x'^2-y'^2=8\). 3. In standard form, \(\frac{x'^2}{8}-\frac{y'^2}{8}=1\). Thus \(a^2=b^2=8\) and \(c^2=a^2+b^2=16\). 4. The principal-coordinate vertices are \((\pm2\sqrt{2}, 0)\). Converting back gives \((2, 2)\) and \((-2, -2)\). 5. The principal-coordinate foci are \((\pm4, 0)\). Converting back gives \((2\sqrt{2}, 2\sqrt{2})\) and \((-2\sqrt{2}, -2\sqrt{2})\).

Principal form: \(\frac{x'^2}{8}-\frac{y'^2}{8}=1\) Vertices: \((2, 2)\), \((-2, -2)\) Foci: \((2\sqrt{2}, 2\sqrt{2})\), \((-2\sqrt{2}, -2\sqrt{2})\)

- The asymptotes come from setting the homogeneous quadratic part equal to zero. - For a centered hyperbola, the principal axes bisect the angles between the asymptotes. - Test the two bisector lines in the original equation to distinguish the transverse direction from the conjugate direction.

1. The asymptotes come from \(3x^2-10xy+3y^2=0\). Setting \(y=mx\) gives \(3m^2-10m+3=0\), so \(m=3\) or \(m=\frac{1}{3}\). 2. The asymptotes are symmetric about the lines \(y=x\) and \(y=-x\), so those angle bisectors give the principal-axis directions. 3. On \(y=x\), the left side of the conic equation becomes \(-4x^2\), which cannot equal \(8\) for real \(x\). On \(y=-x\), it becomes \(16x^2\), so real points occur and this bisector is the transverse-axis direction. 4. Rotate \(45^\circ\) clockwise, using \(x=\frac{x'+y'}{\sqrt{2}}\) and \(y=\frac{-x'+y'}{\sqrt{2}}\). 5. The equation becomes \(8x'^2-2y'^2=8\), or \(x'^2-\frac{y'^2}{4}=1\). 6. The principal vertices are \((\pm1,0)\). Converting back gives \(\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\) and \(\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\).

a) Asymptotes: \(y=3x\) and \(y=\frac{1}{3}x\); their angle bisectors are \(y=x\) and \(y=-x\), the principal-axis directions. b) \(y=-x\) is the transverse-axis direction because the original equation has real points on that bisector, whereas it has none on \(y=x\). c) A valid rotation is \(45^\circ\) clockwise; principal form: \(x'^2-\frac{y'^2}{4}=1\); vertices: \(\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\), \(\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\)

- Treat \(B\) as the mixed-term coefficient in the general quadratic discriminant. - Separate the negative-, positive-, and zero-discriminant ranges before handling the circle subcase. - At the two boundary values, factor the quadratic expression instead of applying a nondegenerate-conic label.

1. Here \(A=C=5\), so the quadratic discriminant is \(B^2-4AC=B^2-100\). 2. If \(|B|<10\), the discriminant is negative. The graph is nonempty, for example because \((0,2)\) satisfies the equation, and the quadratic part is nondegenerate, so the graph is an ellipse. When \(B=0\), the equal pure-square coefficients give the circle \(x^2+y^2=4\); for \(0<|B|<10\), it is a noncircular ellipse. 3. If \(|B|>10\), the discriminant is positive, so the graph is a hyperbola. 4. If \(B=10\), then \(5(x+y)^2=20\), so the graph is the two parallel lines \(x+y=\pm2\). 5. If \(B=-10\), then \(5(x-y)^2=20\), so the graph is the two parallel lines \(x-y=\pm2\).

\(0<|B|<10\): noncircular ellipse \(B=0\): circle \(x^2+y^2=4\) \(|B|>10\): hyperbola \(B=10\): \(x+y=\pm2\) \(B=-10\): \(x-y=\pm2\)

- Equal coefficients on \(x^2\) and \(y^2\) suggest diagonal directions. - Rewrite the quadratic using coordinates along those directions. - Use the semi-axis lengths from standard form to compute area.

1. Because the pure-square coefficients are equal, use \(x'=\frac{x+y}{\sqrt{2}}\) and \(y'=\frac{-x+y}{\sqrt{2}}\). 2. Since \(x^2+y^2=x'^2+y'^2\) and \(xy=\frac{x'^2-y'^2}{2}\), the equation becomes \(10x'^2+4y'^2=40\). 3. The standard form is \(\frac{x'^2}{4}+\frac{y'^2}{10}=1\). 4. Rotation preserves lengths and area, so the ellipse has semi-axis lengths \(2\) and \(\sqrt{10}\), giving area \(2\sqrt{10}\pi\). 5. The minor axis follows the \(x'\)-direction \(y=x\), and the major axis follows the \(y'\)-direction \(y=-x\).

Area: \(2\sqrt{10}\pi\) Major axis: along \(y=-x\) Minor axis: along \(y=x\)

- Substitute a general rotation into the principal-axis equation. - Isolate only the coefficient of the mixed term. - Use the maximum possible magnitude of a sine function.

1. Use \(x'=x\cos\theta+y\sin\theta\) and \(y'=-x\sin\theta+y\cos\theta\). 2. After substitution, the coefficient of \(xy\) is \(2(2-8)\sin\theta\cos\theta=-6\sin 2\theta\). 3. Therefore, its absolute value is \(6|\sin2\theta|\), whose greatest possible value is \(6\). 4. This occurs when \(|\sin2\theta|=1\), so \(2\theta=90^\circ,270^\circ\) modulo \(360^\circ\). 5. In the requested interval, the rotations are \(\theta=45^\circ\) and \(\theta=135^\circ\).

Greatest possible absolute \(xy\)-coefficient: \(6\) Rotation angles: \(45^\circ\) and \(135^\circ\)

- Use the standard formulas for a \(45^\circ\) rotation. - Factor the rotated equation into two linear equations. - Find the angle each line makes with one rotated axis.

1. Use \(x=\frac{x'-y'}{\sqrt{2}}\) and \(y=\frac{x'+y'}{\sqrt{2}}\). 2. Substitution gives \(-2x'^2+4y'^2=0\). 3. Thus, the two lines are \(y'=\pm\frac{x'}{\sqrt{2}}\). 4. These lines make angles \(\pm\arctan\left(\frac{1}{\sqrt{2}}\right)\) with the \(x'\)-axis. 5. Their acute angle is \(2\arctan\left(\frac{1}{\sqrt{2}}\right)\approx70.53^\circ\).

Rotated equation: \(-2x'^2+4y'^2=0\) Lines: \(y'=\pm\frac{x'}{\sqrt{2}}\) Acute angle: \(2\arctan\left(\frac{1}{\sqrt{2}}\right)\approx70.53^\circ\)

- First determine when rotation produces no mixed term. - Separate rotations that preserve each principal axis from rotations that exchange the axes. - Check the order of the two squared-term coefficients.

1. Under a rotation, the mixed-term coefficient is proportional to \((5-3)\sin2\theta\). 2. For the rotated equation to have no mixed term, \(\sin2\theta=0\), so \(\theta=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\). 3. At \(\theta=\frac{\pi}{2}\) or \(\frac{3\pi}{2}\), the axes are exchanged, so the equation becomes \(5x'^2+3y'^2=1\), not the required equation. 4. At \(\theta=0\) or \(\pi\), each principal axis returns to its original line and the equation remains \(3x'^2+5y'^2=1\).

\(\theta=0\) or \(\theta=\pi\)

- Identify \(A\), \(B\), and \(C\) from the quadratic part and use the course rotation formula. - Choose an angle in the first quadrant that satisfies the resulting double-angle equation. - Substitute the rotated-coordinate formulas and verify that the xy-term disappears before classifying.

1. Here \(A=7\), \(B=-6\sqrt{3}\), and \(C=13\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}\): \(\cot(2\theta)=\frac{-6}{-6\sqrt{3}}=\frac{1}{\sqrt{3}}\). 3. Choose \(2\theta=\frac{\pi}{3}\), so \(\theta=\frac{\pi}{6}\). 4. Substitute \(x=u\cos\theta-v\sin\theta\) and \(y=u\sin\theta+v\cos\theta\). 5. With \(\cos\theta=\frac{\sqrt{3}}{2}\) and \(\sin\theta=\frac{1}{2}\), the mixed terms cancel and the equation becomes \(4u^2+16v^2=16\). 6. Dividing by \(16\) gives \(\frac{u^2}{4}+v^2=1\), which is an ellipse.

Rotation angle: \(\theta=\frac{\pi}{6}\) Rotated equation: \(\frac{u^2}{4}+v^2=1\) Conic: ellipse

- Write the equation produced by a \(90^\circ\) rotation. - Subtract the two ellipse equations before solving them simultaneously. - Use the symmetry of the intersection coordinates to identify the quadrilateral.

1. Rotating \(E_1\) by \(90^\circ\) exchanges its axis directions, so \(E_2:\frac{x^2}{9}+\frac{y^2}{25}=1\). 2. Subtracting the two equations gives \(\left(\frac{1}{25}-\frac{1}{9}\right)x^2+\left(\frac{1}{9}-\frac{1}{25}\right)y^2=0\), so \(x^2=y^2\). 3. Substitute \(y^2=x^2\) into either ellipse: \(x^2\left(\frac{1}{25}+\frac{1}{9}\right)=1\). 4. Thus, \(x^2=y^2=\frac{225}{34}\). 5. The four intersections are \(\left(\pm\frac{15}{\sqrt{34}}, \pm\frac{15}{\sqrt{34}}\right)\), with the signs chosen independently. 6. They form an axis-aligned square of side length \(\frac{30}{\sqrt{34}}\), so its area is \(\frac{900}{34}=\frac{450}{17}\).

Intersections: \(\left(\pm\frac{15}{\sqrt{34}}, \pm\frac{15}{\sqrt{34}}\right)\) Square area: \(\frac{450}{17}\) square units

- Identify \(A\), \(B\), and \(C\) and apply \(\cot(2\theta)=\frac{A-C}{B}\). - Choose a convenient angle satisfying the double-angle equation. - Substitute the rotated-coordinate formulas, verify cancellation of the mixed term, and normalize the result.

1. Here \(A=13\), \(B=10\sqrt{3}\), and \(C=3\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}\): \(\cot(2\theta)=\frac{10}{10\sqrt{3}}=\frac{1}{\sqrt{3}}\). 3. Choose \(2\theta=\frac{\pi}{3}\), so \(\theta=\frac{\pi}{6}\). 4. Substitute \(x=x'\cos\theta-y'\sin\theta\) and \(y=x'\sin\theta+y'\cos\theta\). 5. The mixed terms cancel, and the equation becomes \(18x'^2-2y'^2=72\). 6. Dividing by \(72\) gives \(\frac{x'^2}{4}-\frac{y'^2}{36}=1\), so the conic is a hyperbola.

Rotation angle: \(\theta=\frac{\pi}{6}\) Rotated equation: \(\frac{x'^2}{4}-\frac{y'^2}{36}=1\) Conic: hyperbola

- Find the relative angle between the principal axes and the new axes. - Express the principal coordinates using that relative rotation, not either absolute angle alone. - Substitute and simplify the squared expressions.

1. The principal \(x'\)-axis is rotated \(45^\circ\) clockwise relative to the new X-axis because \(30^\circ-75^\circ=-45^\circ\). 2. Therefore, \(x'=\frac{X-Y}{\sqrt{2}}\) and \(y'=\frac{X+Y}{\sqrt{2}}\). 3. Substitute into the principal equation: \(\frac{(X-Y)^2}{32}+\frac{(X+Y)^2}{8}=1\). 4. Multiply by \(32\): \((X-Y)^2+4(X+Y)^2=32\). 5. Expanding gives \(5X^2+6XY+5Y^2=32\).

\(5X^2+6XY+5Y^2=32\)

- Apply the course rotation formula to determine a convenient double angle. - Choose sine and cosine values that satisfy the resulting \(3\)-\(4\)-\(5\) ratio. - After substitution, compare the magnitudes of the two squared-term coefficients and inspect the principal asymptotes.

1. Here \(A=3\), \(B=8\), and \(C=-3\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}=\frac{6}{8}=\frac{3}{4}\). 3. Choose \(\tan(2\theta)=\frac{4}{3}\), which is satisfied by \(\tan\theta=\frac{1}{2}\). Thus, \(\cos\theta=\frac{2}{\sqrt{5}}\) and \(\sin\theta=\frac{1}{\sqrt{5}}\). 4. Use \(x=\frac{2x'-y'}{\sqrt{5}}\) and \(y=\frac{x'+2y'}{\sqrt{5}}\). 5. Substitution cancels the mixed term and gives \(5x'^2-5y'^2=20\). 6. Therefore, the principal equation is \(x'^2-y'^2=4\). 7. Its asymptotes are \(y'=\pm x'\), which are perpendicular, so the hyperbola is rectangular.

One suitable angle: \(\theta=\arctan\left(\frac{1}{2}\right)\) Principal equation: \(x'^2-y'^2=4\) The hyperbola is rectangular.

- Identify \(A\), \(B\), and \(C\) and apply the course formula for \(\cot(2\theta)\). - Select a standard angle whose cotangent matches the resulting value. - Substitute the rotated-coordinate formulas and normalize after confirming that the mixed term cancels.

1. Here \(A=4\), \(B=4\sqrt{3}\), and \(C=8\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}=-\frac{1}{\sqrt{3}}\). 3. Choose \(2\theta=\frac{2\pi}{3}\), so \(\theta=\frac{\pi}{3}\). 4. Substitute \(x=u\cos\theta-v\sin\theta\) and \(y=u\sin\theta+v\cos\theta\). 5. The mixed terms cancel, and the equation becomes \(10u^2+2v^2=20\). 6. Dividing by \(20\) gives \(\frac{u^2}{2}+\frac{v^2}{10}=1\), so the conic is an ellipse.

Rotation angle: \(\theta=\frac{\pi}{3}\) Rotated equation: \(\frac{u^2}{2}+\frac{v^2}{10}=1\) Conic: ellipse

- Apply \(\cot(2\theta)=\frac{A-C}{B}\) to the quadratic coefficients. - Use a half-angle whose sine and cosine come from a simple right triangle. - Substitute the rotated-coordinate formulas and normalize after checking that the mixed term vanishes.

1. Here \(A=13\), \(B=12\), and \(C=-3\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}=\frac{16}{12}=\frac{4}{3}\). 3. Choose \(\tan\theta=\frac{1}{3}\), which gives \(\tan(2\theta)=\frac{3}{4}\). Thus, \(\cos\theta=\frac{3}{\sqrt{10}}\) and \(\sin\theta=\frac{1}{\sqrt{10}}\). 4. Substitute \(x=x'\cos\theta-y'\sin\theta\) and \(y=x'\sin\theta+y'\cos\theta\). 5. The mixed term cancels, and the equation becomes \(15x'^2-5y'^2=60\). 6. Dividing by \(60\) gives \(\frac{x'^2}{4}-\frac{y'^2}{12}=1\), so the conic is a hyperbola.

Rotation angle: \(\theta=\arctan\left(\frac{1}{3}\right)\) Rotated equation: \(\frac{x'^2}{4}-\frac{y'^2}{12}=1\) Conic: hyperbola

- Apply the course formula for \(\cot(2\theta)\) to the three quadratic coefficients. - Select a standard angle matching the resulting cotangent. - Substitute the rotated-coordinate formulas and normalize after the mixed term cancels.

1. Here \(A=11\), \(B=-6\sqrt{3}\), and \(C=5\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}=-\frac{1}{\sqrt{3}}\). 3. Choose \(2\theta=\frac{2\pi}{3}\), so \(\theta=\frac{\pi}{3}\). 4. Substitute \(x=u\cos\theta-v\sin\theta\) and \(y=u\sin\theta+v\cos\theta\). 5. The mixed terms cancel, and the equation becomes \(2u^2+14v^2=28\). 6. Dividing by \(28\) gives \(\frac{u^2}{14}+\frac{v^2}{2}=1\), so the conic is an ellipse.

Rotation angle: \(\theta=\frac{\pi}{3}\) Rotated equation: \(\frac{u^2}{14}+\frac{v^2}{2}=1\) Conic: ellipse

- Apply \(\cot(2\theta)=\frac{A-C}{B}\) even though the original x-squared coefficient is zero. - Use the \(7\)-\(24\)-\(25\) double-angle ratios to obtain simple half-angle values. - Substitute the rotation formulas and normalize after confirming cancellation of the mixed term.

1. Here \(A=0\), \(B=-24\), and \(C=7\). 2. Use \(\cot(2\theta)=\frac{A-C}{B}=\frac{7}{24}\). 3. Choose \(\cos(2\theta)=\frac{7}{25}\) and \(\sin(2\theta)=\frac{24}{25}\). Then \(\cos\theta=\frac{4}{5}\) and \(\sin\theta=\frac{3}{5}\). 4. Substitute \(x=\frac{4}{5}x'-\frac{3}{5}y'\) and \(y=\frac{3}{5}x'+\frac{4}{5}y'\). 5. The mixed term cancels, and the equation becomes \(-9x'^2+16y'^2=144\). 6. Dividing by \(144\) gives \(\frac{y'^2}{9}-\frac{x'^2}{16}=1\), so the conic is a hyperbola.

One suitable angle: \(\theta=\arctan\left(\frac{3}{4}\right)\) Rotated equation: \(\frac{y'^2}{9}-\frac{x'^2}{16}=1\) Conic: hyperbola

- Use the stated rotation to rewrite the mixed quadratic in the new coordinates. - The chord condition is given in the rotated frame, so solve for its endpoints there before converting back. - An orthonormal rotation preserves midpoint relationships and distances.

1. Use \(x=\frac{x'-y'}{\sqrt{2}}\) and \(y=\frac{x'+y'}{\sqrt{2}}\). 2. Then \(x^2+y^2=x'^2+y'^2\) and \(xy=\frac{x'^2-y'^2}{2}\), so the ellipse becomes \(2x'^2+8y'^2=8\), or \(\frac{x'^2}{4}+y'^2=1\). 3. On the chord, \(x'=1\), so \(\frac{1}{4}+y'^2=1\), giving \(y'=\pm\frac{\sqrt{3}}{2}\). 4. Thus the rotated endpoints are \(\left(1,\frac{\sqrt{3}}{2}\right)\) and \(\left(1,-\frac{\sqrt{3}}{2}\right)\). 5. Converting back gives \(\left(\frac{2-\sqrt{3}}{2\sqrt{2}},\frac{2+\sqrt{3}}{2\sqrt{2}}\right)\) and \(\left(\frac{2+\sqrt{3}}{2\sqrt{2}},\frac{2-\sqrt{3}}{2\sqrt{2}}\right)\). 6. Their midpoint is \(\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\). Rotation preserves length, and the y'-coordinates differ by \(\sqrt{3}\), so the chord length is \(\sqrt{3}\).

a) Principal form: \(\frac{x'^2}{4}+y'^2=1\); rotated endpoints: \(\left(1,\frac{\sqrt{3}}{2}\right)\), \(\left(1,-\frac{\sqrt{3}}{2}\right)\) b) Original endpoints: \(\left(\frac{2-\sqrt{3}}{2\sqrt{2}},\frac{2+\sqrt{3}}{2\sqrt{2}}\right)\) and \(\left(\frac{2+\sqrt{3}}{2\sqrt{2}},\frac{2-\sqrt{3}}{2\sqrt{2}}\right)\); midpoint: \(\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\); length: \(\sqrt{3}\)

- Use the same orthonormal rotation on both equations. - Compare what happens to a mixed quadratic with what happens to the rotationally invariant quantity \(x^2+y^2\). - Solve the system in principal coordinates before applying the inverse coordinate transformation.

1. Use \(x=\frac{x'-y'}{\sqrt{2}}\) and \(y=\frac{x'+y'}{\sqrt{2}}\). 2. The ellipse becomes \(x'^2+4y'^2=4\). 3. Because an orthonormal rotation preserves \(x^2+y^2\), the circle becomes \(x'^2+y'^2=2\). 4. Subtracting the circle equation from the ellipse equation gives \(3y'^2=2\), so \(y'=\pm\sqrt{\frac{2}{3}}\). 5. Then \(x'^2=2-y'^2=\frac{4}{3}\), so \(x'=\pm\frac{2}{\sqrt{3}}\), independently of the sign of \(y'\). 6. Write \(x'=\frac{2s}{\sqrt{3}}\) and \(y'=\frac{\sqrt{2}t}{\sqrt{3}}\), where \(s,t\in\{-1,1\}\). Converting back gives \(x=\frac{\sqrt{2}s-t}{\sqrt{3}}\) and \(y=\frac{\sqrt{2}s+t}{\sqrt{3}}\).

a) Ellipse: \(x'^2+4y'^2=4\); circle: \(x'^2+y'^2=2\). The circle is unchanged in form because rotation preserves distance from the origin. b) \(\left(\frac{2s}{\sqrt{3}},\frac{\sqrt{2}t}{\sqrt{3}}\right)\), where \(s,t\in\{-1,1\}\) c) \(\left(\frac{\sqrt{2}s-t}{\sqrt{3}},\frac{\sqrt{2}s+t}{\sqrt{3}}\right)\), where \(s,t\in\{-1,1\}\)

- Express a point on the original x-axis in the ellipse's principal coordinates. - Use the known x-intercept to form an equation involving the rotation angle. - After finding the angle, substitute the orthonormal coordinate rotation into the standard ellipse equation.

1. Along the x-axis, the principal coordinates are \(x'=x\cos\theta\) and \(y'=-x\sin\theta\). 2. Substituting the x-intercept into \(\frac{x'^2}{25}+\frac{y'^2}{9}=1\) gives \(\frac{225}{13}\left(\frac{\cos^2\theta}{25}+\frac{\sin^2\theta}{9}\right)=1\). 3. Therefore, \(\frac{\cos^2\theta}{25}+\frac{\sin^2\theta}{9}=\frac{13}{225}\). 4. Replace \(\sin^2\theta\) by \(1-\cos^2\theta\). After multiplying by \(225\), \(25-16\cos^2\theta=13\), so \(\cos^2\theta=\frac{3}{4}\). 5. Because \(\theta\) is acute, \(\theta=\frac{\pi}{6}\). 6. The principal coordinates are \(x'=\frac{\sqrt{3}x+y}{2}\) and \(y'=\frac{-x+\sqrt{3}y}{2}\). 7. Substitution into the standard equation gives \(\frac{(\sqrt{3}x+y)^2}{100}+\frac{(-x+\sqrt{3}y)^2}{36}=1\), which expands to \(13x^2-8\sqrt{3}xy+21y^2=225\).

\(\theta=\frac{\pi}{6}\) Equation: \(13x^2-8\sqrt{3}xy+21y^2=225\)

- A directrix is perpendicular to the transverse axis. - Compare the directrix's distance from the origin with the principal hyperbola's directrix distance. - Use an orthonormal basis aligned with the directrix normal and its perpendicular direction.

1. The principal hyperbola has \(a=3\), \(b=4\), and \(c=5\). 2. A directrix is perpendicular to the transverse axis and lies at distance \(\frac{a^2}{c}=\frac{9}{5}\) from the center. 3. The line \(3x+4y=9\) has unit normal \(\left(\frac{3}{5}, \frac{4}{5}\right)\) and distance \(\frac{9}{5}\) from the origin, so this unit vector gives the positive \(x'\)-axis direction. 4. Use \(x'=\frac{3x+4y}{5}\) and \(y'=\frac{-4x+3y}{5}\). 5. Substitution into the principal equation gives \(\frac{(3x+4y)^2}{225}-\frac{(-4x+3y)^2}{400}=1\), which simplifies to \(24xy+7y^2=144\). 6. The vertices are \(\pm3\left(\frac{3}{5}, \frac{4}{5}\right)\), and the foci are \(\pm5\left(\frac{3}{5}, \frac{4}{5}\right)\).

Transverse-axis direction: \(\left(\frac{3}{5}, \frac{4}{5}\right)\) Original-coordinate equation: \(24xy+7y^2=144\) Vertices: \(\left(\frac{9}{5}, \frac{12}{5}\right)\), \(\left(-\frac{9}{5}, -\frac{12}{5}\right)\) Foci: \((3, 4)\), \((-3, -4)\)

- Translate each point relative to the focus before applying the physical rotation. - Express the original coordinates in terms of the rotated point coordinates. - Rotate the vertex and directrix geometrically as a check on the transformed equation.

1. A point \((x, y)\) rotated \(90^\circ\) counterclockwise about \((2, 0)\) moves to \((X, Y)=(2-y, x-2)\). 2. The inverse relations are \(x=Y+2\) and \(y=2-X\). 3. Substitute these into the original equation: \((2-X)^2=8(Y+2)\). 4. Therefore the rotated equation is \((X-2)^2=8(Y+2)\). 5. The original vertex \((0, 0)\) rotates to \((2, -2)\), while the rotation center \(F=(2, 0)\) remains the focus. 6. The original directrix \(x=-2\) rotates to \(Y=-4\). The axis is \(X=2\), and the parabola opens upward.

Equation: \((X-2)^2=8(Y+2)\) Vertex: \((2, -2)\) Focus: \((2, 0)\) Directrix: \(Y=-4\) Axis: \(X=2\) Opening direction: upward

- Choose rotated coordinates along the directions \(x+y\) and \(x-y\). - Complete the square only after the mixed term has disappeared. - Convert the center and major-axis displacement back to the original coordinates.

1. Rotate coordinates using \(x'=\frac{x+y}{\sqrt{2}}\) and \(y'=\frac{x-y}{\sqrt{2}}\). 2. The quadratic part becomes \(5x'^2+y'^2\), and the linear part becomes \(-10\sqrt{2}x'\). 3. The equation is \(5x'^2+y'^2-10\sqrt{2}x'+8=0\). 4. Complete the square: \(5(x'-\sqrt{2})^2+y'^2=2\). 5. Thus, \(\frac{(x'-\sqrt{2})^2}{2/5}+\frac{y'^2}{2}=1\), so the conic is an ellipse. 6. The center has \(x'=\sqrt{2}\), \(y'=0\), which converts to \((1, 1)\). 7. The major-axis direction is the \(y'\)-direction \(\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)\), with semi-major axis \(\sqrt{2}\). The minor-axis direction is the \(x'\)-direction \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\), with semi-minor axis \(\sqrt{\frac{2}{5}}\). 8. Moving \(\sqrt{2}\) units from the center along the major direction gives the vertices \((2, 0)\) and \((0, 2)\).

Conic: ellipse Center: \((1, 1)\) Major-axis direction: \(\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)\) Semi-major axis: \(\sqrt{2}\) Minor-axis direction: \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) Semi-minor axis: \(\sqrt{\frac{2}{5}}\) Vertices: \((2, 0)\), \((0, 2)\)

- Express the principal coordinates as orthogonal combinations of \(x\) and \(y\). - Use the inverse rotation to transform the given principal-coordinate point and each principal-axis endpoint. - Find the focal distance in principal coordinates before rotating the two foci.

1. For a \(30^\circ\) rotation of axes, \(x'=\frac{\sqrt{3}x+y}{2}\) and \(y'=\frac{-x+\sqrt{3}y}{2}\). 2. Substitution into the principal equation gives \(\frac{(\sqrt{3}x+y)^2}{144}+\frac{(-x+\sqrt{3}y)^2}{64}=1\). Expanding and simplifying gives \(21x^2-10\sqrt{3}xy+31y^2=576\). 3. The inverse coordinate transformation is \(x=x'\cos30^\circ-y'\sin30^\circ\) and \(y=x'\sin30^\circ+y'\cos30^\circ\). 4. For \(P'=\left(3,2\sqrt{3}\right)\), \(x=\frac{3\sqrt{3}}{2}-\sqrt{3}=\frac{\sqrt{3}}{2}\) and \(y=\frac{3}{2}+3=\frac{9}{2}\). 5. The principal vertices are \((\pm6,0)\). Rotating them gives \((3\sqrt{3},3)\) and \((-3\sqrt{3},-3)\). 6. The principal co-vertices are \((0,\pm4)\). Rotating them gives \((-2,2\sqrt{3})\) and \((2,-2\sqrt{3})\). 7. The focal distance is \(c=\sqrt{36-16}=2\sqrt{5}\). Rotating \((\pm2\sqrt{5},0)\) gives the foci \((\sqrt{15},\sqrt{5})\) and \((-\sqrt{15},-\sqrt{5})\).

a) \(21x^2-10\sqrt{3}xy+31y^2=576\) b) \(P=\left(\frac{\sqrt{3}}{2}, \frac{9}{2}\right)\) c) Vertices: \((3\sqrt{3},3)\) and \((-3\sqrt{3},-3)\); co-vertices: \((-2,2\sqrt{3})\) and \((2,-2\sqrt{3})\); foci: \((\sqrt{15},\sqrt{5})\) and \((-\sqrt{15},-\sqrt{5})\)

- Use the mixed-term coefficients to determine a principal-axis direction for each equation. - Compare the resulting principal coefficients before deciding whether the ellipses are congruent. - A physical rotation changes the directions of both principal axes by the same angle.

1. For \(E_1\), the rotation condition is \(\tan(2\theta)=\frac{B}{A-C}=\frac{-6\sqrt{3}}{7-13}=\sqrt{3}\). Choosing \(\theta=30^\circ\) gives principal coordinates with the \(x'\)-axis directed \(30^\circ\) from the positive x-axis. 2. Under this rotation, \(E_1\) becomes \(4x'^2+16y'^2=100\), or \(\frac{x'^2}{25}+\frac{y'^2}{25/4}=1\). Therefore, its major axis is directed \(30^\circ\) from the positive x-axis. 3. For \(E_2\), the same calculation gives \(\tan(2\theta)=\frac{-6\sqrt{3}}{13-7}=-\sqrt{3}\). Choosing the principal \(x'\)-axis at \(60^\circ\) gives \(4x'^2+16y'^2=100\), or the same principal equation. 4. Thus the major axis of \(E_2\) is directed \(60^\circ\) from the positive x-axis. 5. The acute angle between the major axes is \(60^\circ-30^\circ=30^\circ\). 6. The ellipses have the same center and semi-axis lengths, and rotating \(E_1\) counterclockwise by \(30^\circ\) aligns its major and minor axes with those of \(E_2\). Therefore, this rotation maps \(E_1\) onto \(E_2\).

a) Both have principal equation \(\frac{x'^2}{25}+\frac{y'^2}{25/4}=1\). The major axis of \(E_1\) is directed \(30^\circ\) from the positive x-axis, and the major axis of \(E_2\) is directed \(60^\circ\) from the positive x-axis. b) \(30^\circ\) c) Rotate \(E_1\) counterclockwise by \(30^\circ\) about the origin.

- Locate the center before rotating the axes. - Translate the center to the origin, then use the symmetry of the quadratic coefficients to choose a rotation. - Convert the principal-coordinate asymptotes back through both transformations.

1. The center makes the first-degree terms vanish after translation. Solving \(-10x+26y+62=0\) and \(26x-10y-46=0\) gives the center \((1, -2)\). 2. Let \(X=x-1\) and \(Y=y+2\). The equation becomes \(-5X^2+26XY-5Y^2-72=0\). 3. Rotate \(45^\circ\) with \(x'=\frac{X+Y}{\sqrt{2}}\) and \(y'=\frac{-X+Y}{\sqrt{2}}\). The equation becomes \(8x'^2-18y'^2=72\). 4. The standard form is \(\frac{x'^2}{9}-\frac{y'^2}{4}=1\). 5. In principal coordinates the asymptotes are \(y'=\pm\frac{2}{3}x'\). 6. Converting them back gives \(y=5x-7\) and \(y=\frac{x-11}{5}\).

Center: \((1, -2)\) Principal form: \(\frac{x'^2}{9}-\frac{y'^2}{4}=1\) Asymptotes: \(y=5x-7\) and \(y=\frac{x-11}{5}\)

- Factor the quadratic part before choosing rotated coordinates. - Align one new axis with the repeated linear direction in that square. - Convert the principal focus and directrix back through the rotation.

1. The quadratic part is \(3(x-y)^2\), so use \(x'=\frac{x+y}{\sqrt{2}}\) and \(y'=\frac{-x+y}{\sqrt{2}}\). 2. Since \(x-y=-\sqrt{2}y'\) and \(x+y=\sqrt{2}x'\), the equation becomes \(6y'^2-4\sqrt{2}x'=0\). 3. Thus \(y'^2=\frac{2\sqrt{2}}{3}x'=4px'\), where \(p=\frac{\sqrt{2}}{6}\). The conic is a parabola with vertex at the origin. 4. In principal coordinates, the focus is \((p, 0)\). Converting back gives \((\frac{1}{6}, \frac{1}{6})\). 5. The axis is the \(x'\)-axis, so in original coordinates it is \(y=x\). 6. The directrix is \(x'=-p\), which becomes \(x+y=-\frac{1}{3}\).

Principal form: \(y'^2=\frac{2\sqrt{2}}{3}x'\) Conic: parabola Vertex: \((0, 0)\) Focus: \((\frac{1}{6}, \frac{1}{6})\) Axis: \(y=x\) Directrix: \(x+y=-\frac{1}{3}\)

- Parameterize the ellipse in its principal coordinates. - Rotate the parameterization into the original coordinate system. - Treat each resulting coordinate as a sinusoidal expression and find its amplitude.

1. Parameterize the ellipse by \(x'=5\cos t\) and \(y'=2\sin t\). 2. Since the principal axes are rotated by \(30^\circ\), \(x=5\cos t\cos30^\circ-2\sin t\sin30^\circ\). 3. The greatest possible value of \(|x|\) is the amplitude \(\sqrt{25\cos^2 30^\circ+4\sin^2 30^\circ}=\frac{\sqrt{79}}{2}\). 4. Similarly, \(y=5\cos t\sin30^\circ+2\sin t\cos30^\circ\), so the greatest possible value of \(|y|\) is \(\sqrt{25\sin^2 30^\circ+4\cos^2 30^\circ}=\frac{\sqrt{37}}{2}\). 5. The rectangle extends from \(-\frac{\sqrt{79}}{2}\) to \(\frac{\sqrt{79}}{2}\) horizontally and from \(-\frac{\sqrt{37}}{2}\) to \(\frac{\sqrt{37}}{2}\) vertically. 6. Therefore, its dimensions are \(\sqrt{79}\) by \(\sqrt{37}\).

Smallest axis-aligned rectangle: \(\sqrt{79}\) units by \(\sqrt{37}\) units

- Project the translated position vector onto the major-axis direction and its perpendicular direction. - Use the semi-axis lengths in principal standard form. - Convert focal offsets and directrix positions along the major axis back to the original coordinates.

1. Relative to the center, the principal coordinates are \(x'=\frac{3(x-1)+4(y+2)}{5}=\frac{3x+4y+5}{5}\) and \(y'=\frac{-4(x-1)+3(y+2)}{5}=\frac{-4x+3y+10}{5}\). 2. The ellipse equation is \(\frac{x'^2}{36}+\frac{y'^2}{20}=1\), so in the original coordinates it is \(\frac{(3x+4y+5)^2}{900}+\frac{(-4x+3y+10)^2}{500}=1\). 3. The focal distance is \(c=\sqrt{36-20}=4\). 4. Moving \(4\) units from the center along the major-axis unit vector gives the foci \(\left(\frac{17}{5}, \frac{6}{5}\right)\) and \(\left(-\frac{7}{5}, -\frac{26}{5}\right)\). 5. The eccentricity is \(e=\frac{4}{6}=\frac{2}{3}\), so the directrices are at principal coordinates \(x'=\pm\frac{a}{e}=\pm9\). 6. Converting these lines back gives \(3x+4y=40\) and \(3x+4y=-50\).

Equation: \(\frac{(3x+4y+5)^2}{900}+\frac{(-4x+3y+10)^2}{500}=1\) Foci: \(\left(\frac{17}{5}, \frac{6}{5}\right)\) and \(\left(-\frac{7}{5}, -\frac{26}{5}\right)\) Directrices: \(3x+4y=40\) and \(3x+4y=-50\)

- Read the center and semi-axis lengths from the diagram, then project the displacement from the center onto the two principal directions. - Parameterize the ellipse with those orthonormal directions. - Express the squared distance from the origin as a function of one trigonometric value and compare its endpoint values.

1. From the diagram, the center is \(C=(6, 8)\), the semi-major axis is \(4\), and the semi-minor axis is \(2\). The principal coordinates relative to the center are \(x'=\mathbf u\cdot((x, y)-C)=\frac{3x+4y-50}{5}\) and \(y'=\mathbf v\cdot((x, y)-C)=\frac{-4x+3y}{5}\). 2. The ellipse equation is \(\frac{x'^2}{16}+\frac{y'^2}{4}=1\), so \(\frac{(3x+4y-50)^2}{400}+\frac{(-4x+3y)^2}{100}=1\). 3. Since \(C=10\mathbf u\), parameterize the ellipse as \(R(t)=10\mathbf u+4\mathbf u\cos t+2\mathbf v\sin t\). 4. Because \(\mathbf u\) and \(\mathbf v\) are orthonormal, \(|R(t)|^2=(10+4\cos t)^2+(2\sin t)^2\). 5. Let \(c=\cos t\). Then \(|R(t)|^2=104+80c+12c^2\). For \(-1\le c_1<c_2\le1\), the difference is \((c_2-c_1)(80+12(c_1+c_2))>0\), so this expression increases throughout \([-1,1]\). 6. The minimum occurs at \(c=-1\), giving \(C-4\mathbf u=\left(\frac{18}{5}, \frac{24}{5}\right)\) at distance \(6\). 7. The maximum occurs at \(c=1\), giving \(C+4\mathbf u=\left(\frac{42}{5}, \frac{56}{5}\right)\) at distance \(14\).

a) \(\frac{(3x+4y-50)^2}{400}+\frac{(-4x+3y)^2}{100}=1\) b) Nearest point: \(\left(\frac{18}{5}, \frac{24}{5}\right)\), distance \(6\); farthest point: \(\left(\frac{42}{5}, \frac{56}{5}\right)\), distance \(14\)