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 \(P=(3\sqrt{3}, 3)\): \(u=(3\sqrt{3})(\frac{\sqrt{3}}{2})+3(\frac{1}{2})=6\). 2. Its second coordinate is \(v=-(3\sqrt{3})(\frac{1}{2})+3(\frac{\sqrt{3}}{2})=0\). 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 \(u\)-axis, confirming that the rotation aligns that axis with the ellipse’s major axis.

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

- 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 \(u^2\) and \(v^2\) 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 \(x=\frac{u-v}{\sqrt{2}}\) and \(y=\frac{u+v}{\sqrt{2}}\). 3. Then \(x^2+y^2=u^2+v^2\) and \(xy=\frac{u^2-v^2}{2}\). 4. The equation becomes \(11u^2+5v^2=25\). 5. In standard form, \(\frac{u^2}{25/11}+\frac{v^2}{5}=1\). Both squared terms are positive, so the conic is an ellipse.

A valid rotation is \(\theta=45^\circ\). The principal standard form is \(\frac{u^2}{25/11}+\frac{v^2}{5}=1\), so the conic is an ellipse.

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

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

Principal standard form: \(\frac{u^2}{2}+\frac{v^2}{10/13}=1\) 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 \(u^2\cos^2\theta-2uv\sin\theta\cos\theta+v^2\sin^2\theta+u^2\sin^2\theta+2uv\sin\theta\cos\theta+v^2\cos^2\theta\). 3. The mixed terms cancel, and \(\sin^2\theta+\cos^2\theta=1\), so \(x^2+y^2=u^2+v^2\). 4. Therefore, the rotated equation is \(u^2+v^2=R^2\), 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 \(u^2+v^2=R^2\). 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 \(u=x\cos30^\circ+y\sin30^\circ\) and \(v=-x\sin30^\circ+y\cos30^\circ\). 2. Thus, \(u=\frac{\sqrt{3}}{2}x+\frac{1}{2}y\) and \(v=-\frac{1}{2}x+\frac{\sqrt{3}}{2}y\). 3. Substitute these expressions into \(\frac{u^2}{25}+\frac{v^2}{9}=1\). 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 \(u=\frac{x+y}{\sqrt{2}}\) and \(v=\frac{-x+y}{\sqrt{2}}\). 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 \(x=u\cos\theta-v\sin\theta\) and \(y=u\sin\theta+v\cos\theta\). 5. The equation becomes \(9u^2+v^2=36\), or \(\frac{u^2}{4}+\frac{v^2}{36}=1\). 6. The u-axis is \(30^\circ\) counterclockwise from the x-axis, and the v-axis is perpendicular to it. The major axis is the v-axis.

Rotation angle: \(\theta=\frac{\pi}{6}\) Principal form: \(\frac{u^2}{4}+\frac{v^2}{36}=1\) 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 \(y=-x\)

- 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 \(x=\frac{1}{2}u-\frac{\sqrt{3}}{2}v\) and \(y=\frac{\sqrt{3}}{2}u+\frac{1}{2}v\). After combining like terms, the equation becomes \(16u^2+4v^2=16\). 4. Divide by \(16\) to obtain \(u^2+\frac{v^2}{4}=1\). 5. Both squared terms have positive coefficients, so the conic is an ellipse.

The rotation angle is \(\theta=60^\circ\). In rotated coordinates, the standard form is \(u^2+\frac{v^2}{4}=1\), so the conic is an 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, \(u=x\cos30^\circ+y\sin30^\circ=\frac{\sqrt{3}x+y}{2}\) and \(v=-x\sin30^\circ+y\cos30^\circ=\frac{-x+\sqrt{3}y}{2}\). 2. The principal equation is equivalent to \(4u^2+9v^2=36\). 3. Substitute the expressions for \(u\) and \(v\): \((\sqrt{3}x+y)^2+\frac{9}{4}(-x+\sqrt{3}y)^2=36\). 4. Multiplying by \(4\) and expanding gives \(21x^2-10\sqrt{3}xy+31y^2=144\). 5. Since the larger semi-axis is along the \(u\)-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 \(x=\frac{\sqrt{3}}{2}u+\frac{1}{2}v\) and \(y=-\frac{1}{2}u+\frac{\sqrt{3}}{2}v\). 4. Substitution simplifies the equation to \(2u^2+6v^2=8\), or \(\frac{u^2}{4}+\frac{v^2}{4/3}=1\). 5. The larger denominator is under \(u^2\), so the major axis follows the \(u\)-axis, directed \(30^\circ\) clockwise from the positive x-axis.

One valid rotation is \(\theta=-30^\circ\). Principal form: \(\frac{u^2}{4}+\frac{v^2}{4/3}=1\) The major axis is directed \(30^\circ\) 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 \(x=\frac{u-v}{\sqrt{2}}\) and \(y=\frac{u+v}{\sqrt{2}}\). 2. Then \(xy=\frac{u^2-v^2}{2}\), so the equation becomes \(u^2-v^2=8\). 3. In standard form, \(\frac{u^2}{8}-\frac{v^2}{8}=1\). 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: \(\frac{u^2}{8}-\frac{v^2}{8}=1\) Vertices: \((2, 2)\), \((-2, -2)\) Foci: \((2\sqrt{2}, 2\sqrt{2})\), \((-2\sqrt{2}, -2\sqrt{2})\)

- Find the asymptote slopes from the homogeneous quadratic part. - Principal axes bisect the angles between the asymptotes. - Test the two bisector directions to identify the transverse axis before rotating.

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 angle-bisector directions of these asymptotes are \(y=x\) and \(y=-x\). Substitution into the original conic shows the transverse axis is along \(y=-x\). 3. Rotate \(45^\circ\) clockwise, using \(x=\frac{u+v}{\sqrt{2}}\) and \(y=\frac{-u+v}{\sqrt{2}}\). 4. The equation becomes \(8u^2-2v^2=8\), or \(u^2-\frac{v^2}{4}=1\). 5. The principal vertices are \((\pm1, 0)\). Converting back gives \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})\) and \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\).

Asymptotes: \(y=3x\) and \(y=\frac{1}{3}x\) A valid rotation is \(45^\circ\) clockwise. Principal form: \(u^2-\frac{v^2}{4}=1\) Vertices: \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})\), \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\)

- Equal pure-square coefficients make a \(45^\circ\) rotation especially useful. - Classify the graph from the signs of the two rotated squared-term coefficients. - Treat a zero rotated coefficient as a separate degenerate case.

1. Because the coefficients of \(x^2\) and \(y^2\) are equal, rotate \(45^\circ\) using \(u=\frac{x+y}{\sqrt{2}}\) and \(v=\frac{-x+y}{\sqrt{2}}\). 2. The equation becomes \((5+\frac{B}{2})u^2+(5-\frac{B}{2})v^2=20\). 3. If \(0<|B|<10\), both coefficients are positive but unequal, so the graph is a noncircular ellipse. When \(B=0\), the coefficients are equal and the graph is the circle \(x^2+y^2=4\). 4. If \(|B|>10\), the coefficients have opposite signs, so the graph is a hyperbola. 5. If \(B=10\), the equation is \(10u^2=20\), giving the parallel lines \(x+y=2\) and \(x+y=-2\). 6. If \(B=-10\), the equation is \(10v^2=20\), giving the parallel lines \(-x+y=2\) and \(-x+y=-2\).

\(0<|B|<10\): 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 \(u=\frac{x+y}{\sqrt{2}}\) and \(v=\frac{-x+y}{\sqrt{2}}\). 2. Since \(x^2+y^2=u^2+v^2\) and \(xy=\frac{u^2-v^2}{2}\), the equation becomes \(10u^2+4v^2=40\). 3. The standard form is \(\frac{u^2}{4}+\frac{v^2}{10}=1\). 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 \(u\)-direction \(y=x\), and the major axis follows the \(v\)-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 \(u=x\cos\theta+y\sin\theta\) and \(v=-x\sin\theta+y\cos\theta\). 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 \(x=\frac{u-v}{\sqrt{2}}\) and \(y=\frac{u+v}{\sqrt{2}}\). 2. Substitution gives \(-2u^2+4v^2=0\). 3. Thus, the two lines are \(v=\pm\frac{u}{\sqrt{2}}\). 4. These lines make angles \(\pm\arctan\left(\frac{1}{\sqrt{2}}\right)\) with the \(u\)-axis. 5. Their acute angle is \(2\arctan\left(\frac{1}{\sqrt{2}}\right)\approx70.53^\circ\).

Rotated equation: \(-2u^2+4v^2=0\) Lines: \(v=\pm\frac{u}{\sqrt{2}}\) 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 \(5u^2+3v^2=1\), not the required equation. 4. At \(\theta=0\) or \(\pi\), each principal axis returns to its original line and the equation remains \(3u^2+5v^2=1\).

\(\theta=0\) 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 \(x=u\cos\theta-v\sin\theta\) and \(y=u\sin\theta+v\cos\theta\). 5. The mixed terms cancel, and the equation becomes \(18u^2-2v^2=72\). 6. Dividing by \(72\) gives \(\frac{u^2}{4}-\frac{v^2}{36}=1\), so the conic is a hyperbola.

Rotation angle: \(\theta=\frac{\pi}{6}\) Rotated equation: \(\frac{u^2}{4}-\frac{v^2}{36}=1\) 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 \(u\)-axis is rotated \(45^\circ\) clockwise relative to the new \(X\)-axis because \(30^\circ-75^\circ=-45^\circ\). 2. Therefore, \(u=\frac{X-Y}{\sqrt{2}}\) and \(v=\frac{X+Y}{\sqrt{2}}\). 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 \(x=\frac{2u-v}{\sqrt{5}}\) and \(y=\frac{u+2v}{\sqrt{5}}\). 5. Substitution cancels the mixed term and gives \(5u^2-5v^2=20\). 6. Therefore, the principal equation is \(u^2-v^2=4\). 7. Its asymptotes are \(v=\pm u\), which are perpendicular, so the hyperbola is rectangular.

One suitable angle: \(\theta=\arctan\left(\frac{1}{2}\right)\) Principal equation: \(u^2-v^2=4\) 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 \(x=u\cos\theta-v\sin\theta\) and \(y=u\sin\theta+v\cos\theta\). 5. The mixed term cancels, and the equation becomes \(15u^2-5v^2=60\). 6. Dividing by \(60\) gives \(\frac{u^2}{4}-\frac{v^2}{12}=1\), so the conic is a hyperbola.

Rotation angle: \(\theta=\arctan\left(\frac{1}{3}\right)\) Rotated equation: \(\frac{u^2}{4}-\frac{v^2}{12}=1\) 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 \(x=\frac{4}{5}u-\frac{3}{5}v\) and \(y=\frac{3}{5}u+\frac{4}{5}v\). 5. The mixed term cancels, and the equation becomes \(-9u^2+16v^2=144\). 6. Dividing by \(144\) gives \(\frac{v^2}{9}-\frac{u^2}{16}=1\), so the conic is a hyperbola.

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

- Choose rotated coordinates that align with the quadratic form's principal directions. - Rewrite both the conic and the chord line in the same rotated coordinate system. - Use the fact that an orthonormal rotation preserves distances.

1. Use \(x=\frac{u-v}{\sqrt2}\) and \(y=\frac{u+v}{\sqrt2}\). 2. Then \(x^2+y^2=u^2+v^2\) and \(xy=\frac{u^2-v^2}{2}\), so the ellipse becomes \(2u^2+8v^2=8\), or \(\frac{u^2}{4}+v^2=1\). 3. The chord line becomes \(\sqrt2u=\sqrt2\), so \(u=1\). 4. Substitution gives \(\frac14+v^2=1\), so \(v=\pm\frac{\sqrt3}{2}\). 5. Converting back gives the endpoints \(\left(\frac{2-\sqrt3}{2\sqrt2}, \frac{2+\sqrt3}{2\sqrt2}\right)\) and \(\left(\frac{2+\sqrt3}{2\sqrt2}, \frac{2-\sqrt3}{2\sqrt2}\right)\). 6. Their midpoint is \(\left(\frac1{\sqrt2}, \frac1{\sqrt2}\right)\). Since rotation preserves length and the v-coordinates differ by \(\sqrt3\), the chord length is \(\sqrt3\).

Endpoints: \(\left(\frac{2-\sqrt3}{2\sqrt2}, \frac{2+\sqrt3}{2\sqrt2}\right)\) and \(\left(\frac{2+\sqrt3}{2\sqrt2}, \frac{2-\sqrt3}{2\sqrt2}\right)\) Midpoint: \(\left(\frac1{\sqrt2}, \frac1{\sqrt2}\right)\) Chord length: \(\sqrt3\)

- Choose a rotation that diagonalizes the quadratic expression containing the mixed term. - A circle centered at the origin keeps the same equation under an orthonormal rotation. - Solve the two equations first in the rotated coordinates, then convert every sign combination back.

1. Rotate the axes \(45^\circ\) counterclockwise using \(x=\frac{u-v}{\sqrt2}\) and \(y=\frac{u+v}{\sqrt2}\). 2. The ellipse becomes \(2u^2+8v^2=8\), or \(u^2+4v^2=4\). 3. The circle is unchanged by rotation, so it becomes \(u^2+v^2=2\). 4. Subtracting the circle equation from the ellipse equation gives \(3v^2=2\), so \(v=\pm\sqrt{\frac23}\). 5. Then \(u^2=2-v^2=\frac43\), so \(u=\pm\frac{2}{\sqrt3}\). 6. Write \(u=\frac{2s}{\sqrt3}\) and \(v=\frac{\sqrt2t}{\sqrt3}\), where \(s, t\in\{-1,1\}\). Converting back gives \(x=\frac{\sqrt2s-t}{\sqrt3}\) and \(y=\frac{\sqrt2s+t}{\sqrt3}\).

The four intersection points are \(\left(\frac{\sqrt2s-t}{\sqrt3}, \frac{\sqrt2s+t}{\sqrt3}\right)\), 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 \(u=x\cos\theta\) and \(v=-x\sin\theta\). 2. Substituting the x-intercept into \(\frac{u^2}{25}+\frac{v^2}{9}=1\) 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 \(\cos^2\theta=\frac34\). 5. Because \(\theta\) is acute, \(\theta=\frac{\pi}{6}\). 6. The principal coordinates are \(u=\frac{\sqrt3x+y}{2}\) and \(v=\frac{-x+\sqrt3y}{2}\). 7. Substitution into the standard equation gives \(\frac{(\sqrt3x+y)^2}{100}+\frac{(-x+\sqrt3y)^2}{36}=1\), which expands to \(13x^2-8\sqrt3xy+21y^2=225\).

\(\theta=\frac{\pi}{6}\) Equation: \(13x^2-8\sqrt3xy+21y^2=225\)

- 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 \(\frac{a^2}{c}=\frac95\) from the center. 3. The line \(3x+4y=9\) has unit normal \(\left(\frac35, \frac45\right)\) and distance \(\frac95\) from the origin, so this unit vector gives the positive \(u\)-axis direction. 4. Use \(u=\frac{3x+4y}{5}\) and \(v=\frac{-4x+3y}{5}\). 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 \(\pm3\left(\frac35, \frac45\right)\), and the foci are \(\pm5\left(\frac35, \frac45\right)\).

Transverse-axis direction: \(\left(\frac35, \frac45\right)\) Original-coordinate equation: \(24xy+7y^2=144\) Vertices: \(\left(\frac95, \frac{12}{5}\right)\), \(\left(-\frac95, -\frac{12}{5}\right)\) 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 \(u=\frac{x+y}{\sqrt2}\) and \(v=\frac{x-y}{\sqrt2}\). 2. The quadratic part becomes \(5u^2+v^2\), and the linear part becomes \(-10\sqrt2u\). 3. The equation is \(5u^2+v^2-10\sqrt2u+8=0\). 4. Complete the square: \(5(u-\sqrt2)^2+v^2=2\). 5. Thus \(\frac{(u-\sqrt2)^2}{2/5}+\frac{v^2}{2}=1\), so the conic is an ellipse. 6. The center has \(u=\sqrt2\), \(v=0\), which converts to \((1, 1)\). 7. The major-axis direction is the \(v\)-direction \(\left(\frac1{\sqrt2}, -\frac1{\sqrt2}\right)\), with semi-major axis \(\sqrt2\). The minor-axis direction is the \(u\)-direction \(\left(\frac1{\sqrt2}, \frac1{\sqrt2}\right)\), with semi-minor axis \(\sqrt{\frac25}\). 8. Moving \(\sqrt2\) units from the center along the major direction gives the vertices \((2, 0)\) and \((0, 2)\).

Conic: ellipse Center: \((1, 1)\) Major-axis direction: \(\left(\frac1{\sqrt2}, -\frac1{\sqrt2}\right)\) Semi-major axis: \(\sqrt2\) Minor-axis direction: \(\left(\frac1{\sqrt2}, \frac1{\sqrt2}\right)\) Semi-minor axis: \(\sqrt{\frac25}\) Vertices: \((2, 0)\), \((0, 2)\)

- Express the principal coordinates as orthogonal combinations of \(x\) and \(y\). - Use the inverse rotation to transform the given 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, \(u=\frac{\sqrt3x+y}{2}\) and \(v=\frac{-x+\sqrt3y}{2}\). 2. Substitution into the principal equation gives \(\frac{(\sqrt3x+y)^2}{144}+\frac{(-x+\sqrt3y)^2}{64}=1\). Expanding and simplifying gives \(21x^2-10\sqrt3xy+31y^2=576\). 3. The inverse coordinate transformation is \(x=u\cos30^\circ-v\sin30^\circ\) and \(y=u\sin30^\circ+v\cos30^\circ\). 4. For \(P'=\left(3,2\sqrt3\right)\), \(x=\frac{3\sqrt3}{2}-\sqrt3=\frac{\sqrt3}{2}\) and \(y=\frac32+3=\frac92\). 5. The principal vertices are \((\pm6,0)\). Rotating them gives \((3\sqrt3,3)\) and \((-3\sqrt3,-3)\). 6. The principal co-vertices are \((0,\pm4)\). Rotating them gives \((-2,2\sqrt3)\) and \((2,-2\sqrt3)\). 7. The focal distance is \(c=\sqrt{36-16}=2\sqrt5\). Rotating \((\pm2\sqrt5,0)\) gives the foci \((\sqrt{15},\sqrt5)\) and \((-\sqrt{15},-\sqrt5)\).

a) \(21x^2-10\sqrt3xy+31y^2=576\) b) \(P=\left(\frac{\sqrt3}{2}, \frac92\right)\) c) Vertices: \((3\sqrt3,3)\) and \((-3\sqrt3,-3)\); co-vertices: \((-2,2\sqrt3)\) and \((2,-2\sqrt3)\); foci: \((\sqrt{15},\sqrt5)\) and \((-\sqrt{15},-\sqrt5)\)

- 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 \(\tan(2\theta)=\frac{B}{A-C}=\frac{-6\sqrt3}{7-13}=\sqrt3\). Choosing \(\theta=30^\circ\) gives principal coordinates with the u-axis directed \(30^\circ\) from the positive x-axis. 2. Under this rotation, \(E_1\) becomes \(4u^2+16v^2=100\), or \(\frac{u^2}{25}+\frac{v^2}{\frac{25}{4}}=1\). Therefore, its major axis is directed \(30^\circ\) from the positive x-axis. 3. For \(E_2\), the same calculation gives \(\tan(2\theta)=\frac{-6\sqrt3}{13-7}=-\sqrt3\). Choosing the principal u-axis at \(60^\circ\) gives \(4u^2+16v^2=100\), 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 \(\frac{u^2}{25}+\frac{v^2}{\frac{25}{4}}=1\). 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 \(u=\frac{X+Y}{\sqrt{2}}\) and \(v=\frac{-X+Y}{\sqrt{2}}\). The equation becomes \(8u^2-18v^2=72\). 4. The standard form is \(\frac{u^2}{9}-\frac{v^2}{4}=1\). 5. In principal coordinates the asymptotes are \(v=\pm\frac{2}{3}u\). 6. Converting them back gives \(y=5x-7\) and \(y=\frac{x-11}{5}\).

Center: \((1, -2)\) Principal form: \(\frac{u^2}{9}-\frac{v^2}{4}=1\) 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 \(u=\frac{x+y}{\sqrt{2}}\) and \(v=\frac{-x+y}{\sqrt{2}}\). 2. Since \(x-y=-\sqrt{2}v\) and \(x+y=\sqrt{2}u\), the equation becomes \(6v^2-4\sqrt{2}u=0\). 3. Thus \(v^2=\frac{2\sqrt{2}}{3}u=4pu\), 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 u-axis, so in original coordinates it is \(y=x\). 6. The directrix is \(u=-p\), which becomes \(x+y=-\frac{1}{3}\).

Principal form: \(v^2=\frac{2\sqrt{2}}{3}u\) 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 \(u=5\cos t\) and \(v=2\sin t\). 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 \(u=\frac{3(x-1)+4(y+2)}{5}=\frac{3x+4y+5}{5}\) and \(v=\frac{-4(x-1)+3(y+2)}{5}=\frac{-4x+3y+10}{5}\). 2. The ellipse equation is \(\frac{u^2}{36}+\frac{v^2}{20}=1\), 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 \(u=\pm\frac{a}{e}=\pm9\). 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 \(3x+4y=-50\)

- 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. The principal coordinates relative to the center are \(U=\mathbf u\cdot((x, y)-C)=\frac{3x+4y-50}{5}\) and \(V=\mathbf v\cdot((x, y)-C)=\frac{-4x+3y}{5}\). 2. The ellipse equation is \(\frac{U^2}{16}+\frac{V^2}{4}=1\), 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 \((c_2-c_1)\bigl(80+12(c_1+c_2)\bigr)>0\), 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 \(14\)