- Group the x-terms and y-terms before completing the squares. - Add each completing-square value to both sides of the equation. - Equal coefficients on the two squared terms in standard form identify a circle.

1. Group the variable terms: \((x^2+8x)+(y^2-6y)=-9\). 2. Complete both squares by adding \(16\) and \(9\) to both sides. 3. This gives \((x+4)^2+(y-3)^2=16\). 4. The graph is a circle centered at \((-4, 3)\) with radius \(4\).

The conic is a circle with standard form \((x+4)^2+(y-3)^2=16\), center \((-4, 3)\), and radius \(4\).

- Compare the signs between the two squared terms. - You do not need to complete any squares because the equation is already in standard form.

1. The equation is already in standard form. 2. The two squared terms are subtracted, which is the standard-form pattern for a hyperbola.

Hyperbola

- Notice that all four choices have the same vertex and opening direction, so those features alone cannot decide the answer. - Use the second marked point as additional information about the parabola's width. - Think about what coefficient would make that point satisfy the vertex-form equation.

1. The marked vertex is \(V=(1,-2)\), so each choice has the correct vertex and opening direction. 2. Use the second point \(P=(2,-1)\). In \(y+2=a(x-1)^2\), substituting \(P\) gives \(1=a\), so \(a=1\). 3. The matching equation is \(y+2=(x-1)^2\).

B. \(y+2=(x-1)^2\)

- Compare the signs in front of the squared terms. - Check whether one square is missing entirely. - Equal positive square coefficients give a special case.

1. In a), both squared terms are positive with unequal coefficients, so the conic is an ellipse. 2. In b), the squared terms have opposite signs, so it is a hyperbola. 3. In c), only one variable is squared, so it is a parabola. 4. In d), the squared terms have equal positive coefficients, so it is a circle.

a) Ellipse b) Hyperbola c) Parabola d) Circle

- How small can a sum of real squares be? - A product equals zero when at least one factor is zero. - You do not need to complete a square here.

1. In a), a sum of two squares is zero only when both variables are zero, giving only \((0,0)\). 2. In b), the sum of two real squares cannot be negative, so there are no real points. 3. In c), factor \((x-y)(x+y)=0\). The graph is the two lines \(y=x\) and \(y=-x\).

a) One point: \((0,0)\) b) No real points c) Two lines: \(y=x\) and \(y=-x\)

- Choose two clearly readable points on the same branch with different x-coordinates. - Test all three expressions at those points, not just whether they are positive. - Then check whether the expression behaves the same way on the other branch.

1. The graph passes through \((1,4)\) and \((2,2)\). Their \(x^2\) values differ, and their \(y^2\) values differ, but both have \(xy=4\). 2. On the lower-left branch both coordinates are negative, so their product remains \(4\). 3. The equation is \(xy=4\), a rectangular hyperbola with the coordinate axes as asymptotes.

\(xy\) stays equal to \(4\). The graph is a rectangular hyperbola.

- Group the x-terms and y-terms before completing the square. - After completing the square, move the constant so the right side is positive, then divide to make the right side \(1\). - For an ellipse, the larger denominator identifies the direction of the major axis.

1. Group the x-terms and y-terms: \(4(x^2-4x)+9(y^2+2y)-11=0\). 2. Complete the square in each group: \(4((x-2)^2-4)+9((y+1)^2-1)-11=0\). 3. Simplify to get \(4(x-2)^2+9(y+1)^2=36\). 4. Divide by \(36\): \(\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1\). 5. Both squared terms are positive, so the graph is an ellipse centered at \((2, -1)\). Since \(a=3\) along the horizontal major axis, the vertices are \((-1, -1)\) and \((5, -1)\).

The conic is an ellipse with standard form \(\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1\). Its center is \((2, -1)\), and its vertices are \((-1, -1)\) and \((5, -1)\).

- Group the x-terms and y-terms, keeping the negative coefficient attached to the entire y-group. - Complete each square before moving the constant to the other side. - In standard form, opposite signs identify a hyperbola, and the positive term gives the transverse-axis direction.

1. Group the x-terms and y-terms: \(25(x^2+6x)-16(y^2-4y)-239=0\). 2. Complete the square in each group: \(25((x+3)^2-9)-16((y-2)^2-4)-239=0\). 3. Simplify to obtain \(25(x+3)^2-16(y-2)^2=400\). 4. Divide by \(400\): \(\frac{(x+3)^2}{16}-\frac{(y-2)^2}{25}=1\). 5. The opposite signs show that the graph is a horizontal hyperbola centered at \((-3, 2)\). Since \(a=4\), its vertices are \((-7, 2)\) and \((1, 2)\).

The conic is a horizontal hyperbola with standard form \(\frac{(x+3)^2}{16}-\frac{(y-2)^2}{25}=1\). Its center is \((-3, 2)\), and its vertices are \((-7, 2)\) and \((1, 2)\).

- Interpret each square root as a distance from the variable point to a fixed point. - Identify which conic is defined by a constant sum of two focal distances. - Use the distance sum and focal separation to determine the semi-axis lengths.

1. The equation states that the sum of the distances from \((x, y)\) to the fixed points \((4, 0)\) and \((-4, 0)\) is \(12\). 2. A constant sum of distances to two distinct foci defines an ellipse. 3. The foci give \(c=4\), and the constant distance sum is \(2a=12\), so \(a=6\). 4. Then \(b^2=a^2-c^2=36-16=20\). 5. The foci lie on the x-axis and the center is the origin, so the standard form is \(\frac{x^2}{36}+\frac{y^2}{20}=1\).

The conic is an ellipse with standard form \(\frac{x^2}{36}+\frac{y^2}{20}=1\).

- Group the terms involving each variable before completing squares. - Watch how the parameter is already paired with the x-terms. - After reaching standard form, separate the changing feature from the fixed feature.

1. Complete the square in \(x\): \(x^2-2kx+k^2=(x-k)^2\). 2. Complete the square in \(y\): \(y^2+4y=(y+2)^2-4\). 3. The equation becomes \((x-k)^2+(y+2)^2=4\). 4. For every real \(k\), the graph is a circle with center \((k, -2)\) and radius \(2\). 5. As \(k\) varies, the center moves along the horizontal line \(y=-2\).

For every real \(k\), the conic is a circle of radius \(2\) centered at \((k, -2)\). The centers trace the line \(y=-2\).

- Isolate the terms involving \(x\) before completing the square. - Add the same value to both sides when you complete the square. - A standard form with only one squared variable represents a parabola.

1. Move the y-term and constant: \(x^2-6x=8y-1\). 2. Complete the square in \(x\): \(x^2-6x+9=8y+8\). 3. Rewrite the equation as \((x-3)^2=8(y+1)\). 4. This is a vertical parabola with vertex \((3, -1)\).

The conic is a parabola with standard form \((x-3)^2=8(y+1)\) and vertex \((3, -1)\).

- Group the x-terms and y-terms before completing each square. - Divide by the positive constant so the right side becomes \(1\). - For an ellipse, the larger denominator identifies the major-axis direction.

1. Group the variable terms: \(9(x^2-2x)+4(y^2+2y)-23=0\). 2. Complete both squares: \(9((x-1)^2-1)+4((y+1)^2-1)-23=0\). 3. Simplify to \(9(x-1)^2+4(y+1)^2=36\). 4. Divide by \(36\): \(\frac{(x-1)^2}{4}+\frac{(y+1)^2}{9}=1\). 5. The graph is an ellipse centered at \((1, -1)\). Its major axis is vertical, so the vertices are \((1, 2)\) and \((1, -4)\).

The conic is an ellipse with standard form \(\frac{(x-1)^2}{4}+\frac{(y+1)^2}{9}=1\). Its center is \((1, -1)\), and its vertices are \((1, 2)\) and \((1, -4)\).

- Keep the negative coefficient attached to the entire y-group while completing the square. - Opposite signs in standard form identify a hyperbola, and the positive term gives its opening direction. - Use the denominator square roots to form the asymptote slopes.

1. Group the variable terms: \(4(x^2+4x)-9(y^2-2y)-29=0\). 2. Complete both squares: \(4((x+2)^2-4)-9((y-1)^2-1)-29=0\). 3. Simplify to \(4(x+2)^2-9(y-1)^2=36\). 4. Divide by \(36\): \(\frac{(x+2)^2}{9}-\frac{(y-1)^2}{4}=1\). 5. The graph is a horizontal hyperbola centered at \((-2, 1)\), with vertices \((-5, 1)\) and \((1, 1)\). 6. Its asymptotes are \(y-1=\pm\frac{2}{3}(x+2)\).

The conic is a horizontal hyperbola with standard form \(\frac{(x+2)^2}{9}-\frac{(y-1)^2}{4}=1\). Center: \((-2, 1)\) Vertices: \((-5, 1)\), \((1, 1)\) Asymptotes: \(y-1=\pm\frac{2}{3}(x+2)\)

- Keep the negative sign attached to the full y-group when completing the square. - The positive squared term in standard form gives the transverse-axis direction. - Form each asymptote by replacing the right side \(1\) with \(0\) and solving for \(y\).

1. Group the variable terms: \(9(x^2-4x)-4(y^2+2y)-4=0\). 2. Complete both squares: \(9((x-2)^2-4)-4((y+1)^2-1)-4=0\). 3. Simplify to \(9(x-2)^2-4(y+1)^2=36\). 4. Divide by \(36\): \(\frac{(x-2)^2}{4}-\frac{(y+1)^2}{9}=1\). 5. The conic is a horizontal hyperbola centered at \((2,-1)\), with vertices \((0,-1)\) and \((4,-1)\). 6. Its asymptotes are \(y+1=\pm\frac{3}{2}(x-2)\).

The conic is a horizontal hyperbola with standard form \(\frac{(x-2)^2}{4}-\frac{(y+1)^2}{9}=1\). Center: \((2,-1)\) Vertices: \((0,-1)\), \((4,-1)\) Asymptotes: \(y+1=\pm\frac{3}{2}(x-2)\)

- Group the terms in each variable before completing the squares. - Divide by the positive constant so the right side becomes \(1\). - The larger denominator identifies the ellipse’s major-axis direction.

1. Group the variable terms: \(25(x^2-4x)+9(y^2+6y)-44=0\). 2. Complete both squares: \(25((x-2)^2-4)+9((y+3)^2-9)-44=0\). 3. Simplify to \(25(x-2)^2+9(y+3)^2=225\). 4. Divide by \(225\): \(\frac{(x-2)^2}{9}+\frac{(y+3)^2}{25}=1\). 5. The graph is an ellipse centered at \((2, -3)\). Its major axis is vertical, so the vertices are \((2, 2)\) and \((2, -8)\).

The conic is an ellipse with standard form \(\frac{(x-2)^2}{9}+\frac{(y+3)^2}{25}=1\). Its center is \((2, -3)\), and its vertices are \((2, 2)\) and \((2, -8)\).

- Look for two binomial factors whose product reproduces the mixed and linear terms. - A constant product of two shifted variables describes a familiar conic. - The zero lines of the two factors reveal important geometric features.

1. Group the product and linear terms by completing a product: \(xy-2x+3y-10=(x+3)(y-2)-4\). 2. Therefore, the equation is \((x+3)(y-2)=4\). 3. This is a translated rectangular hyperbola with center \((-3, 2)\). 4. Its asymptotes occur when either factor is zero: \(x=-3\) and \(y=2\).

The conic is a rectangular hyperbola. Rewritten equation: \((x+3)(y-2)=4\) Center: \((-3, 2)\) Asymptotes: \(x=-3\) and \(y=2\)

- Write each squared distance directly from the distance formula without taking square roots. - Expand the two expressions together and look for cancellation of linear terms. - Compare the simplified result with standard conic forms.

1. The squared distances are \(PA^2=(x+2)^2+y^2\) and \(PB^2=(x-2)^2+y^2\). 2. The condition is \((x+2)^2+y^2+(x-2)^2+y^2=20\). 3. Expanding and combining terms gives \(2x^2+2y^2+8=20\). 4. Therefore, \(x^2+y^2=6\). 5. This is a circle centered at \((0, 0)\) with radius \(\sqrt{6}\).

Conic: circle Standard equation: \(x^2+y^2=6\) Center: \((0, 0)\) Radius: \(\sqrt{6}\)

- Isolate the y-terms before completing the square. - Add the same completing-square value to both sides. - The sign of the coefficient multiplying \(x-h\) determines the horizontal opening direction.

1. Move the x-term and constant: \(y^2-10y=-8x-1\). 2. Complete the square by adding \(25\) to both sides: \(y^2-10y+25=-8x+24\). 3. Rewrite as \((y-5)^2=-8(x-3)\). 4. The graph is a parabola with vertex \((3, 5)\). Since the coefficient is negative, it opens to the left.

The conic is a parabola with standard form \((y-5)^2=-8(x-3)\), vertex \((3, 5)\), and opening direction to the left.

- Group the x-terms and y-terms before completing the square in each variable. - After completing the squares, move the constant so the right side is positive and divide to make it \(1\). - For an ellipse, the larger denominator identifies the major-axis direction.

1. Group the x-terms and y-terms: \(16(x^2+4x)+9(y^2-6y)+1=0\). 2. Complete each square: \(16((x+2)^2-4)+9((y-3)^2-9)+1=0\). 3. Simplify to obtain \(16(x+2)^2+9(y-3)^2=144\). 4. Divide by \(144\): \(\frac{(x+2)^2}{9}+\frac{(y-3)^2}{16}=1\). 5. Both squared terms are positive and the right side is \(1\), so the graph is an ellipse centered at \((-2,3)\). 6. The larger denominator is under the y-term, so the major axis is vertical.

Conic: ellipse Standard form: \(\frac{(x+2)^2}{9}+\frac{(y-3)^2}{16}=1\) Center: \((-2, 3)\) Major axis: vertical

- Group the x-terms and y-terms, keeping each leading coefficient outside its group. - Complete the square in both variables and move the constant to the other side. - Opposite signs identify a hyperbola; the positive squared term gives its opening direction.

1. Group the variable terms: \(9(x^2-6x)-4(y^2+4y)+29=0\). 2. Complete the squares: \(9((x-3)^2-9)-4((y+2)^2-4)+29=0\). 3. Simplify: \(9(x-3)^2-4(y+2)^2=36\). 4. Divide by \(36\): \(\frac{(x-3)^2}{4}-\frac{(y+2)^2}{9}=1\). 5. The squared terms have opposite signs, so the conic is a hyperbola centered at \((3,-2)\). 6. The positive x-term shows that it opens left and right.

Conic: hyperbola Standard form: \(\frac{(x-3)^2}{4}-\frac{(y+2)^2}{9}=1\) Center: \((3, -2)\) Opening direction: left and right

- Treat the square root as a nonnegative radial quantity. - Solve the resulting equation in that radial quantity without dividing by it. - Translate every radial solution back into a Cartesian graph.

1. Let \(r=\sqrt{x^2+y^2}\), where \(r\ge0\). 2. The equation becomes \(r^2=2r\), so \(r(r-2)=0\). 3. Therefore, either \(r=0\) or \(r=2\). 4. The case \(r=0\) gives the isolated point \((0, 0)\). 5. The case \(r=2\) gives the circle \(x^2+y^2=4\). 6. The union of a circle and an isolated point is not a single conic.

Complete graph: the circle \(x^2+y^2=4\) together with the isolated point \((0, 0)\) It is not a single conic.

- Translate the distance ratio into an equation before simplifying. - Square both sides and collect the quadratic and linear terms. - Complete the square to identify the conic.

1. The distance condition is \(\sqrt{(x-3)^2+y^2}=2\sqrt{(x+1)^2+y^2}\). 2. Squaring gives \((x-3)^2+y^2=4[(x+1)^2+y^2]\). 3. Expanding and simplifying yields \(3x^2+14x+3y^2-5=0\). 4. Divide by \(3\) and complete the square: \(\left(x+\frac{7}{3}\right)^2+y^2=\frac{64}{9}\). 5. The conic is a circle centered at \(\left(-\frac{7}{3}, 0\right)\) with radius \(\frac{8}{3}\).

Conic: circle Standard equation: \(\left(x+\frac{7}{3}\right)^2+y^2=\frac{64}{9}\) Center: \(\left(-\frac{7}{3}, 0\right)\) Radius: \(\frac{8}{3}\)

- Isolate the x-quadratic expression and complete its square. - A quadratic equation with only one squared variable represents a parabola when it has a real graph. - Compare the result with \((x-h)^2=4p(y-k)\) to read the vertex and opening direction.

1. Group the x-terms: \(x^2-8x=-6y-10\). 2. Complete the square by adding \(16\) to both sides: \((x-4)^2=-6y+6\). 3. Factor the right side: \((x-4)^2=-6(y-1)\). 4. Only one variable is squared, so the conic is a parabola. 5. Its vertex is \((4,1)\), and the negative coefficient shows that it opens downward.

Conic: parabola Standard form: \((x-4)^2=-6(y-1)\) Vertex: \((4, 1)\) Opening direction: downward

- Record the sign restriction imposed by the principal square root. - Use squaring only together with the original domain conditions. - Compare the resulting subset with the complete circle.

1. The square root requires \(25-x^2\ge0\), so \(-5\le x\le5\). 2. It also requires \(y\ge0\). 3. Squaring gives \(x^2+y^2=25\), but only the points with \(y\ge0\) satisfy the original equation. 4. Therefore, the graph is the upper semicircle of radius \(5\) centered at the origin, including its endpoints. 5. A semicircle is only part of a conic, not a complete conic graph.

The graph is the upper semicircle \(x^2+y^2=25\) with \(y\ge0\). It is not a complete conic.

- Group the x-terms and y-terms before completing both squares. - Equal positive coefficients on the squared terms indicate a circle after normalization. - Read the center from the translated squares and the radius from the positive right side.

1. Divide by \(4\) or group directly: \(4(x^2+4x)+4(y^2-6y)+3=0\). 2. Complete the squares: \(4((x+2)^2-4)+4((y-3)^2-9)+3=0\). 3. Simplify to obtain \(4(x+2)^2+4(y-3)^2=49\). 4. Divide by \(4\): \((x+2)^2+(y-3)^2=\frac{49}{4}\). 5. The conic is a circle centered at \((-2,3)\) with radius \(\frac{7}{2}\).

Conic: circle Standard form: \((x+2)^2+(y-3)^2=\frac{49}{4}\) Center: \((-2, 3)\) Radius: \(\frac{7}{2}\)

- Group the x-terms and y-terms before completing each square. - Divide by the positive constant so the right side becomes \(1\). - The denominators are the squares of the semi-axis lengths, not the full axis lengths.

1. Group the variable terms: \(4(x^2-8x)+25(y^2+4y)=-64\). 2. Complete the squares: \(4\bigl((x-4)^2-16\bigr)+25\bigl((y+2)^2-4\bigr)=-64\). 3. Simplify: \(4(x-4)^2+25(y+2)^2=100\). 4. Divide by \(100\): \(\frac{(x-4)^2}{25}+\frac{(y+2)^2}{4}=1\). 5. The conic is an ellipse centered at \((4,-2)\), with major-axis length \(10\) and minor-axis length \(4\).

Conic: ellipse Standard form: \(\frac{(x-4)^2}{25}+\frac{(y+2)^2}{4}=1\) Center: \((4, -2)\) Axis lengths: \(10\) and \(4\)

- Express the midpoint coordinates in terms of the endpoint coordinates. - Rewrite the endpoint constraint using the midpoint variables. - Identify the resulting second-degree equation from its product form.

1. Let the midpoint be \((x, y)\). Then \(x=\frac{a}{2}\) and \(y=\frac{b}{2}\). 2. Therefore \(a=2x\) and \(b=2y\). 3. Substituting into \(ab=24\) gives \((2x)(2y)=24\), so \(xy=6\). 4. This is a rectangular hyperbola centered at the origin with asymptotes \(x=0\) and \(y=0\).

The locus is \(xy=6\), a rectangular hyperbola with asymptotes the coordinate axes.

- Group each variable's quadratic and linear terms before completing the squares. - Opposite signs on the normalized squared terms identify a hyperbola. - Use the positive term for the opening direction and translate \(y=\pm\frac{b}{a}x\) to the center.

1. Group the variable terms: \(16(x^2+4x)-25(y^2-6y)=561\). 2. Complete the squares: \(16\bigl((x+2)^2-4\bigr)-25\bigl((y-3)^2-9\bigr)=561\). 3. Simplify: \(16(x+2)^2-25(y-3)^2=400\). 4. Divide by \(400\): \(\frac{(x+2)^2}{25}-\frac{(y-3)^2}{16}=1\). 5. The graph is a horizontal hyperbola centered at \((-2,3)\), with vertices \((-7,3)\) and \((3,3)\). 6. Its asymptotes are \(y-3=\pm\frac45(x+2)\).

Conic: hyperbola Standard form: \(\frac{(x+2)^2}{25}-\frac{(y-3)^2}{16}=1\) Center: \((-2, 3)\) Vertices: \((-7, 3)\) and \((3, 3)\) Asymptotes: \(y-3=\pm\frac45(x+2)\)

- Factor the coefficient of \(x^2\) from the x-terms before completing the square. - Add and subtract the square of half the coefficient of \(x\) inside the parentheses. - Compare the result with the standard form of a vertical parabola.

1. Group the x-terms: \(9(x^2-6x)-4y+65=0\). 2. Complete the square inside the parentheses: \(9\bigl((x-3)^2-9\bigr)-4y+65=0\). 3. Simplify: \(9(x-3)^2-4y-16=0\). 4. Solve for the squared expression: \((x-3)^2=\frac{4}{9}(y+4)\). 5. This matches \((x-h)^2=4p(y-k)\), so the graph is a parabola with vertex \((3, -4)\). Because \(4p=\frac49>0\), it opens upward.

The standard form is \((x-3)^2=\frac{4}{9}(y+4)\). The conic is a parabola with vertex \((3, -4)\), and it opens upward.

- Group the x-terms and y-terms before completing either square. - Add the square of half each linear coefficient to both sides. - Compare the result with \((x-h)^2+(y-k)^2=r^2\).

1. Group the variable terms and move the constant: \((x^2-8x)+(y^2+6y)=11\). 2. Complete the square in each group: \((x-4)^2-16+(y+3)^2-9=11\). 3. Add the constants to the right side: \((x-4)^2+(y+3)^2=36\). 4. This is a circle centered at \((4, -3)\) with radius \(6\).

Standard form: \((x-4)^2+(y+3)^2=36\) Conic: circle Center: \((4, -3)\) Radius: \(6\)

- Factor the leading coefficient from each variable group before completing the square. - After completing both squares, divide so the right side is \(1\). - The larger denominator identifies the major-axis direction.

1. Group the x-terms and y-terms: \(16(x^2-4x)+9(y^2+2y)-71=0\). 2. Complete both squares: \(16\bigl((x-2)^2-4\bigr)+9\bigl((y+1)^2-1\bigr)-71=0\). 3. Simplify: \(16(x-2)^2+9(y+1)^2=144\). 4. Divide by \(144\): \(\frac{(x-2)^2}{9}+\frac{(y+1)^2}{16}=1\). 5. The graph is a vertical ellipse centered at \((2, -1)\), with semi-major axis \(4\) and semi-minor axis \(3\). 6. The vertices are \((2, -1\pm4)\), or \((2, 3)\) and \((2, -5)\). 7. The co-vertices are \((2\pm3, -1)\), or \((5, -1)\) and \((-1, -1)\).

Standard form: \(\frac{(x-2)^2}{9}+\frac{(y+1)^2}{16}=1\) Conic: vertical ellipse Center: \((2, -1)\) Vertices: \((2, 3)\) and \((2, -5)\) Co-vertices: \((5, -1)\) and \((-1, -1)\)

- Expand the squared expression and look for a cancellation. - Complete the square in the remaining squared variable. - Compare the result with a translated parabola equation.

1. Expand the right side: \(x^2+y^2-4x=y^2-6y+9\). 2. The \(y^2\) terms cancel, leaving \(x^2-4x+6y-9=0\). 3. Complete the square in \(x\): \((x-2)^2=-6\left(y-\frac{13}{6}\right)\). 4. This is a vertical parabola with \(4p=-6\), so \(p=-\frac32\). 5. The vertex is \(\left(2, \frac{13}{6}\right)\). The focus is \(\left(2, \frac{13}{6}-\frac32\right)=\left(2, \frac23\right)\). 6. The directrix is \(y=\frac{13}{6}+\frac32=\frac{11}{3}\), and the parabola opens downward along the axis \(x=2\).

Conic: parabola Standard form: \((x-2)^2=-6\left(y-\frac{13}{6}\right)\) Vertex: \(\left(2, \frac{13}{6}\right)\) Focus: \(\left(2, \frac23\right)\) Directrix: \(y=\frac{11}{3}\) Axis: \(x=2\) Opening direction: downward

- Ask what distinguishes a circle from a noncircular ellipse in standard form. - Rewrite the equation with \(1\) on the right. - Compare the two denominators after normalization.

1. Positive squared terms are consistent with an ellipse, including a circle as a special case, so their signs alone do not establish that the conic is a circle. 2. Divide by \(16\): \(\frac{(x-2)^2}{16}+\frac{(y+1)^2}{4}=1\). 3. The squared terms have different denominators, so the semi-axis lengths are different. 4. Therefore the conic is an ellipse, not a circle.

The conic is an ellipse. Aroha's mistake is treating same-sign squared terms as sufficient for a circle; a circle also requires equal squared-term coefficients after the equation is normalized.

- Look for a product of factors in one equation versus a sum of squares in the other. - In each equation, identify the common point where both offsets are zero. - The hyperbola approaches the lines where the product factors vanish.

1. The product of two different linear factors equal to a nonzero constant gives a rectangular hyperbola. Its center is \((2,-1)\), and its asymptotes are \(x=2\) and \(y=-1\). 2. The sum of two squared offsets equal to a positive constant gives a circle centered at \((2,-1)\) with radius \(\sqrt{3}\).

First: rectangular hyperbola; center \((2,-1)\); asymptotes \(x=2\), \(y=-1\) Second: circle; center \((2,-1)\)

- In each equation, identify the coefficient of \(xy\), even though it is not a squared term. - Compare the result of the stated expression in the two cases. - Think about which coefficient changes the outcome despite the positive squared terms.

1. In the first equation, \(A=2\), \(B=3\), and \(C=3\). Then \(B^2-4AC=9-24=-15<0\), so it represents an ellipse. 2. In the second equation, \(A=1\), \(B=4\), and \(C=2\). Then \(B^2-4AC=16-8=8>0\), so it represents a hyperbola. 3. Both equations have positive coefficients on the two squared terms. The mixed \(xy\)-term changes the type, so the two square-term signs alone do not decide it.

\(2x^2+3xy+3y^2=1\): \(B^2-4AC=-15\), ellipse. \(x^2+4xy+2y^2=1\): \(B^2-4AC=8\), hyperbola. The \(xy\)-coefficient matters as well as the squared-term coefficients.

- Expand each equation to identify its quadratic coefficients. - Think about how an equation can describe a graph without drawing it. - Check whether every solution of each equation lies on one curved graph.

1. Both equations have quadratic coefficients \(A=1\), \(B=0\), and \(C=0\), so \(B^2-4AC=0\). 2. The first is an upward-opening parabola with vertex \((2,-1)\). 3. The second gives \(x-2=3\) or \(x-2=-3\), so its graph is the pair of vertical lines \(x=5\) and \(x=-1\). 4. A zero discriminant alone does not rule out a degenerate conic.

Both equations have \(A=1\), \(B=0\), \(C=0\), so \(B^2-4AC=0\). \((x-2)^2=4(y+1)\) is a parabola with vertex \((2,-1)\). \((x-2)^2=9\) is the pair of lines \(x=-1\) and \(x=5\). The student's claim is false: \(B^2-4AC=0\) can also describe a degenerate case.

- Identify the coefficients of the three quadratic terms in each equation. - Use the sign of the quadratic discriminant to separate the main conic families. - Check the equal-coefficient case separately for a circle.

1. For a general quadratic, compare \(B^2-4AC\), where \(A\), \(B\), and \(C\) are the coefficients of \(x^2\), \(xy\), and \(y^2\). 2. For a), \(B^2-4AC=4^2-4(3)(3)=-20<0\). The mixed term rules out a circle, so the graph is an ellipse. 3. For b), \(B^2-4AC=4^2-4(1)(1)=12>0\), so the graph is a hyperbola. 4. For c), \(B^2-4AC=12^2-4(4)(9)=0\). Also, the equation is \((2x+3y)^2+x-y=0\), and the linear term varies along the quadratic form's zero direction \((3, -2)\), so the graph is a nondegenerate parabola. 5. For d), division by \(5\) gives \(x^2+y^2=4\), so the graph is a circle.

a) ellipse b) hyperbola c) parabola d) circle

- 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 \(B=10\): \(x+y=\pm2\) \(B=-10\): \(x-y=\pm2\)

- Use the given distance ratio to identify eccentricity before expanding. - Write the point-to-line distance with an absolute value. - Complete the square after removing the radical.

1. The distance ratio is \(e=2>1\), so the locus is a hyperbola. 2. The defining equation is \(\sqrt{x^2+y^2}=2|x-3|\). 3. Squaring gives \(x^2+y^2=4(x-3)^2\). 4. Simplifying and completing the square gives \(3(x-4)^2-y^2=12\), or \(\frac{(x-4)^2}{4}-\frac{y^2}{12}=1\). 5. The center is \((4, 0)\), with \(a=2\), \(b^2=12\), and \(c=4\). 6. The vertices are \((2, 0)\) and \((6, 0)\). Since one focus is \((0, 0)\), the second focus is \((8, 0)\).

Hyperbola: \(\frac{(x-4)^2}{4}-\frac{y^2}{12}=1\) Center: \((4, 0)\) Vertices: \((2, 0)\), \((6, 0)\) Second focus: \((8, 0)\)

- Expand the products and compare the result with standard conic forms. - Complete the square in the variable with a linear term. - Interpret the original sum of products as a dot product.

1. Expand the equation: \(x^2-2x-15+y^2-1=0\). 2. Complete the square in x: \((x-1)^2+y^2=17\). 3. Therefore, the conic is a circle centered at \((1, 0)\) with radius \(\sqrt{17}\). 4. The original expression is the dot product \((x+3, y-1)\cdot(x-5, y+1)\). 5. These vectors run from \((-3, 1)\) and \((5, -1)\) to \((x, y)\). Their dot product is zero exactly when the two segments are perpendicular. 6. Thus, the circle is also the locus of points that form a right angle with the fixed segment joining \((-3, 1)\) and \((5, -1)\) as the hypotenuse.

Conic: circle Standard form: \((x-1)^2+y^2=17\) Center: \((1, 0)\) Radius: \(\sqrt{17}\) Geometrically, the equation requires the segments from \((x, y)\) to \((-3, 1)\) and \((5, -1)\) to be perpendicular.

- Substitute the two signed-distance expressions before classifying the locus. - Expand and complete squares to reveal the center. - Compare the final asymptotes with the two lines used in the distance condition.

1. Substitute the signed-distance expressions: \(\frac{(x+y)(x-y-4)}{2}=2\). 2. Therefore \((x+y)(x-y-4)=4\). 3. Expanding gives \(x^2-y^2-4x-4y=4\). 4. Complete the squares: \((x-2)^2-(y+2)^2=4\). 5. The standard form is \(\frac{(x-2)^2}{4}-\frac{(y+2)^2}{4}=1\). 6. This is a rectangular hyperbola centered at \((2, -2)\), with vertices \((0, -2)\) and \((4, -2)\). 7. Its asymptotes are \(y+2=\pm(x-2)\), which are exactly the two original lines \(L_1\) and \(L_2\).

Conic: rectangular hyperbola Standard form: \(\frac{(x-2)^2}{4}-\frac{(y+2)^2}{4}=1\) Center: \((2, -2)\) Vertices: \((0, -2)\), \((4, -2)\) Asymptotes: \(x+y=0\), \(x-y-4=0\)

- Identify the two fixed points in the distance expressions. - Compare their separation with the required sum of distances. - Use a basic geometric inequality before attempting algebraic squaring.

1. The two fixed points are \((5,0)\) and \((-5,0)\), which are \(10\) units apart. 2. For every point \((x,y)\), the triangle inequality gives \(\sqrt{(x-5)^2+y^2}+\sqrt{(x+5)^2+y^2}\ge10\). 3. The equation requires this sum to equal \(8\), which is smaller than the distance between the two fixed points. 4. Therefore, no real point satisfies the equation. The real graph is empty. 5. A distance-sum equation represents a real ellipse only when the constant is greater than the distance between the two foci. Riku's classification ignores this existence condition.

The real graph is empty. Riku is not correct: the required distance sum \(8\) is less than the \(10\)-unit distance between the two fixed points, so no real ellipse exists.