Conic Sections - Pre-Calculus

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(x-2)^2}{100}-\frac{(y-6)^2}{36}=1$

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $m=\pm\frac{6}{10}=\pm\frac{3}{5}$ .

Now, use the point-slope form of a line in addition to the center of the hyperbola to find the equations of the asymptotes.

The center is at .

The first equation of the asymptote would be the following:

$y-6=\frac{3}{5}(x-2)$

$y-6=\frac{3}{5}x-\frac{6}{5}$

$y=\frac{3}{5}x+\frac{24}{5}$

The second equation of the asymptote would be the following:

$y-6=-\frac{3}{5}(x-2)$

$y-6=-\frac{3}{5}x+\frac{6}{5}$

$y=-\frac{3}{5}+\frac{36}{5}$

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(x-5)^2}{24}-\frac{(y+3)^2}{25}=1$

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, $a=\sqrt{24}=2\sqrt6$ and .

Thus, the slopes for its asymptotes are $m=\pm\frac{5}{2\sqrt6}=\pm\frac{5\sqrt6}{12}$ .

Now, use the point-slope form of a line in addition to the center of the hyperbola to find the equations of the asymptotes.

The center is at .

To find the equations of the asymptotes, use the point-slope form of a line.

$y+3=\pm\frac{5\sqrt6}{12}(x-5)$

$y=\frac{5\sqrt6}{12}(x-5)-3$

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and terms together.

Factor out from the terms and from the terms.

Complete the squares. Remember to add the amount amount to both sides of the equation!

Subtract from both sides of the equation:

Divide both sides by .

$\frac{x^2-18x+81}{4}-\frac{y^2+2y+1}{9}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\frac{(x-9)^2}{4}-\frac{(y+1)^2}{9}=1$

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $m=\pm\frac{3}{2}$ .

Plug in the center of the hyperbola into the point-slope form of a line to find the equations of the asymptotes. The center of the hyperbola is .

$y+1=\pm\frac{3}{2}(x-9)$

Now, simplify each equation. For the first equation,

$y+1=\frac{3}{2}(x-9)$

$y+1=\frac{3}{2}x-\frac{27}{2}$

$y=\frac{3}{2}x-\frac{29}{2}$

For the second equation,

$y+1=-\frac{3}{2}(x-9)$

$y+1=-\frac{3}{2}x+\frac{27}{2}$

$y=-\frac{3}{2}x-\frac{25}{2}$

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(x-12)^2}{144}-\frac{(y+\sqrt3)^2}{169}=1$

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $m=\pm\frac{13}{12}$

Now, plug in the coordinates for the center of the hyperbola $(12, \sqrt3)$ into the point-slope form of the line to find the equations of the asymptotes.

$y-\sqrt3=\pm\frac{13}{12}(x-12)$

$y=\pm\frac{13}{12}(x-12)+\sqrt3$

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and terms together.

Factor out from the terms and from the terms.

Complete the squares. Remember to add the amount amount to both sides of the equation!

Add to both sides of the equation:

Divide both sides by .

$\frac{x^2+6x+9}{25}-\frac{y^2-22y+121}{4}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\frac{(x+3)^2}{25}-\frac{(y-11)^2}{4}=1$

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $\pm\frac{2}{5}$ .

Now, use the center of the hyperbola, , to plug into the point-slope form of a line to find the equations of the asymptotes.

The first asymptote has the following equation:

$y-11=\frac{2}{5}(x+3)$

$y-11=\frac{2}{5}x+\frac{6}{5}$

$y=\frac{2}{5}x+\frac{61}{5}$

The second asymptote has the following equation:

$y-11=-\frac{2}{5}(x+3)$

$y-11=-\frac{2}{5}x-\frac{6}{5}$

$y=-\frac{2}{5}x+\frac{49}{5}$

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{y^2}{9}-\frac{x^2}{25}=1$

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and .

Since the center is , the equations for its asymptotes are $y=\pm\frac{3}{5}x$ .

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(y-9)^2}{169}-\frac{(x-2)^2}{225}=1$

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $\pm\frac{13}{15}$ .

Now, plug in the center of the hyperbola, into the point-slope form of a line to find the equations of the asymptotes.

For the first asymptote,

$y-9=\frac{13}{15}(x-2)$

$y-9=\frac{13}{15}x-\frac{26}{15}$

$y=\frac{13}{15}x+\frac{109}{15}$

For the second asymptote,

$y-9=-\frac{13}{15}(x-2)$

$y-9=-\frac{13}{15}x+\frac{26}{15}$

$y=-\frac{13}{15}x+\frac{161}{15}$

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{(x-5)^2}{25}+\frac{(y-1)^2}{36}=1$

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is horizontal and the endpoints are and .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{(x-1)^2}{64}+\frac{(y-3)^2}{25}=1$

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is vertical and the endpoints are and .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

$25x^2-50x+64y^2-384y-999=0\textup{}$

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides of the equation!

Add to both sides.

Divide by on both sides.

$\frac{x^2-2x+1}{64}+\frac{y^2-6y+9}{25}=1$

Now factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x-1)^2}{64}+\frac{(y-3)^2}{25}=1$

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is vertical and the endpoints are and .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides of the equation!

Subtract from both sides.

Divide by on both sides.

$\frac{x^2+8x+16}{20}+\frac{y^2-6x+9}{25}=1$

Now factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x+4)^2}{20}+\frac{(y-3)^2}{25}=1$

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, $a=\sqrt{20}=2\sqrt5$ and . Since , the minor axis is horizontal and the endpoints are $(-4+2\sqrt5, 3)$ and $(-4-2\sqrt5, 3)$ .

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Question

Find the equations of the asymptotes of the hyperbola with the following equation:

$\frac{x^2}{36}-\frac{y^2}{49}=1$

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where is the center of the hyperbola.

The equations of the asymptotes for this hyperbola are given by the following equations:

$y=\pm\frac{b}{a}x$

For the hyperbola in question, and .

Thus, the equations for its asymptotes are $y=\pm\frac{7}{6}x$

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

$\frac{(x-2)^2}{225}+\frac{(y+11)^2}{256}=1$

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

When , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, and . Since , the major axis is vertical and the endpoints are and .

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

$\frac{(x+5)^2}{169}+\frac{(y-3)^2}{144}=1$

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

When , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, and . Since , the major axis is horizontal and the endpoints are and

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

Tap to reveal answer

Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides of the equation!

Add to both sides.

Divide by on both sides.

$\frac{x^2-8x+16}{49}+\frac{y^2-6y+9}{25}=1$

Now factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x-4)^2}{49}+\frac{(y-3)^2}{25}=1$

Recall that when , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, and . Since , the major axis is horizontal and the endpoints are and .

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides of the equation!

Subtract from both sides.

Divide by on both sides.

$\frac{x^2-12x+36}{18}+\frac{y^2-20x+100}{9}=1$

Now factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x-6)^2}{18}+\frac{(y-10)^2}{9}=1$

Recall that when , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, $a=\sqrt{18}=3\sqrt2$ and . Since , the major axis is horizontal and the endpoints are $(6+3\sqrt2, 10)$ and $(6-3\sqrt2, 10)$ .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{x^2}{100}+\frac{y^2}{169}=1$

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is horizontal and the endpoints are and

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{x^2}{225}+\frac{y^2}{400}=1$

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Answer

Recall the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is horizontal and the endpoints are and

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Question

Which of the following could NOT be a possible number of solutions of a system of quadratic and linear equations?

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Answer

Recall that the solution of a system of equations is given by the intersection points of the graphs. Thus, this question is really asking how many times a parabola and a line can intersect. Visualize a parabola and a line. For this purpose, let's say that the parabola is facing up. If a line is drawn horizontally under the vertex, then it would not intersect the parabola at all, so the system would have no solutions. If the line is tangent to, or just skims the edge, of the parabola then it would only intersect once and the system would have one solution. If the line goes straight through the parabola, then it would intersect twice. There is no other option for the orientation of the line and parabola. Thus, there cannot be three solutions.

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Question

Find the coordinate of intersection, if possible: and .

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Answer

To solve for x and y, set both equations equal to each other and solve for x.

Substitute into either parabola.

The coordinate of intersection is .

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