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Precalculus : Understand features of hyperbolas and ellipses

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Example Questions

Example Question #71 : Understand Features Of Hyperbolas And Ellipses

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

9x^2-4y^2-162x-8y+689=0

Possible Answers:

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

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

$y=\frac{9}{4}x-\frac{14}{4}$

$y=-\frac{9}{4}x+\frac{3}{4}$

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

$y=-\frac{1}{9}x+\frac{16}{9}$

$y=\frac{2}{3}x+\frac{4}{3}$

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

Correct answer:

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

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

Explanation:

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 (h,k) 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.

9x^2-4y^2-162x-8y+689=0

9x^2-162x-4y^2-8y+689=0

Factor out from the terms and from the terms.

9(x^2-18x)-4(y^2+2y)+689=0

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

9(x^2-18x+81)-4(y^2+2y+1)+689=725

Subtract 689 from both sides of the equation:

9(x^2-18x+81)-4(y^2+2y+1)=36

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, a=2 and b=3 .

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 (9, -1) .

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

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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Example Question #72 : Understand Features Of Hyperbolas And Ellipses

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

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

Possible Answers:

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

$y=\pm\frac{144}{169}(x-12)-\sqrt3$

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

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

Correct answer:

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

Explanation:

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 (h,k) 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=12 and b=13 .

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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Example Question #1742 : Pre Calculus

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

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

Possible Answers:

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

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

$y=\frac{5}{3}x-\frac{7}{3}$

$y=-\frac{5}{3}x+\frac{8}{3}$

$y=\frac{25}{9}x-\frac{8}{9}$

$y=-\frac{25}{9}x+\frac{2}{3}$

$y=\frac{9}{25}x-\frac{1}{5}$

$y=-\frac{9}{25}x+\frac{9}{5}$

Correct answer:

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

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

Explanation:

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 (h,k) 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 b=6 .

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 (2, 6) .

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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Example Question #73 : Understand Features Of Hyperbolas And Ellipses

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

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

Possible Answers:

$y=\pm\frac{3}{5}x$

$y=\pm\frac{25}{9}x$

$y=\pm\frac{9}{25}x$

$y=\pm\frac{5}{3}x$

Correct answer:

$y=\pm\frac{3}{5}x$

Explanation:

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 (h,k) 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=5 and b=3 .

Since the center is (0, 0) , the equations for its asymptotes are $y=\pm\frac{3}{5}x$ .

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Example Question #1744 : Pre Calculus

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

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

Possible Answers:

y=2x-9

y=-2x+8

$y=\frac{15}{13}x-\frac{20}{13}$

$y=-\frac{15}{13}x+\frac{5}{13}$

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

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

$y=\frac{2}{9}x-\frac{7}{2}$

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

Correct answer:

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

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

Explanation:

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 (h,k) 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=15 and b=13 .

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

Now, plug in the center of the hyperbola, (2, 9) 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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Example Question #1745 : Pre Calculus

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

$\frac{(y-1)^2}{8}-\frac{(x+2)^2}{36}=1$

Possible Answers:

$y=\pm\frac{3\sqrt2}{2}(x-2)-1$

$y=\pm\frac{\sqrt2}{3}(x+2)+1$

$y=\pm\frac{1}{8}(x-2)-1$

$y=\pm\frac{2\sqrt2}{3}(x-1)+9$

Correct answer:

$y=\pm\frac{\sqrt2}{3}(x+2)+1$

Explanation:

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 (h,k) is the center of the hyperbola.

The slopes of this hyperbola are given by the following:

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

For the hyperbola in question, a=6 and $b=\sqrt8=2\sqrt2$ .

Thus, the slopes for its asymptotes are $\pm\frac{2\sqrt2}{6}=\pm\frac{\sqrt2}{3}$ .

Now, plug in the center of the hyperbola into the point-slope form of the equation fo the line to get the equations of the asymptotes.

The center of the hyperbola is (-2, 1) . The equations of the asymptotes are then:

$y-1=\pm\frac{\sqrt2}{3}(x+2)$

$y=\pm\frac{\sqrt2}{3}(x+2)+1$

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Example Question #71 : Hyperbolas And Ellipses

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

$\frac{(y-4)^2}{196}-\frac{(x+\pi)^2}{400}=1$

Possible Answers:

$y=\pm\frac{13}{20}(x-\pi)+8$

$y=\pm\frac{13}{20}(x+\pi)+4$

$y=\pm\frac{\pi}{4}(x-2)+1$

$y=\pm\frac{20}{13}(x+4)+\pi$

Correct answer:

$y=\pm\frac{13}{20}(x+\pi)+4$

Explanation:

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 (h,k) is the center of the hyperbola.

The slopes of this hyperbola are given by the following:

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

For the hyperbola in question, a=20 and b=13 .

Thus, the slopes for its asymptotes are $\pm\frac{13}{20}$ .

Now, plug in the center of the hyperbola into the point-slope form of an equation of a line to find the equations of the asymptotes. The center of the hyperbola is $(-\pi, 4)$ .

The equations of the asymptotes can be then given by the following:

$y-4=\pm\frac{13}{20}(x+\pi)$

$y=\pm\frac{13}{20}(x+\pi)+4$

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Example Question #71 : Understand Features Of Hyperbolas And Ellipses

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

36y^2-16x^2-144y-32x-448=0

Possible Answers:

$y=\frac{3}{2}x+3$

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

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

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

$y=\pm\frac{4}{3}x$

$y=\frac{2}{3}x+\frac{8}{3}$

$y=-\frac{2}{3}x+\frac{4}{3}$

Correct answer:

$y=\frac{2}{3}x+\frac{8}{3}$

$y=-\frac{2}{3}x+\frac{4}{3}$

Explanation:

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 (h,k) 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.

36y^2-16x^2-144y-32x-448=0

36y^2-144y-16x^2-32x-448=0

Factor out from the terms and from the terms.

36(y^2-4y)-16(x^2+2x)-448=0

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

36(y^2-4y+4)-16(x^2+2x+1)-448=98

Add 448 to both sides of the equation:

36(y^2-4y+4)-16(x^2+2x+1)=576

Divide both sides by 576 .

$\frac{y^2-4y+4}{16}-\frac{x^2+2x+1}{36}=1$

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

$\frac{(y-2)^2}{16}-\frac{(x+1)^2}{36}=1$

The slopes of this hyperbola are given by the following:

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

For the hyperbola in question, a=6 and b=4 .

Thus, the slopes for its asymptotes are $\pm\frac{4}{6}=\pm\frac{2}{3}$ .

Now, plug in the center of the hyperbola, (-1, 2) into the point-slope form of the equation of a line to get the equations of the asymptotes.

For the first equation,

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

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

$y=\frac{2}{3}x+\frac{8}{3}$

For the second equation,

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

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

$y=-\frac{2}{3}x+\frac{4}{3}$

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Example Question #78 : Understand Features Of Hyperbolas And Ellipses

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

4y^2-54x^2-24y+108x-234=0

Possible Answers:

$y=\pm\frac{4\sqrt3}{2}(x+3)-1$

$y=\pm\frac{1}{3}x$

$y=\pm\frac{2\sqrt6}{3}(x-1)+3$

$y=\pm\frac{3\sqrt6}{2}(x-1)+3$

Correct answer:

$y=\pm\frac{3\sqrt6}{2}(x-1)+3$

Explanation:

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 (h,k) 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.

4y^2-54x^2-24y+108x-234=0

4y^2-24y-54x^2+108x-234=0

Factor out from the terms and from the terms.

4(y^2-6y)-54(x^2-2x)-234=0

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

4(y^2-6y+9)-54(x^2-2x+1)-234=-18

Add 234 to both sides of the equation:

4(y^2-6y+9)-54(x^2-2x+1)=216

Divide both sides by 216 .

$\frac{y^2-6y+9}{54}-\frac{x^2-2x+1}{4}=1$

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

$\frac{(y-3)^2}{54}-\frac{(x-1)^2}{4}=1$

The slopes of this hyperbola are given by the following:

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

For the hyperbola in question, a=2 and $b=\sqrt{54}=3\sqrt6$ .

Thus, the slopes for its asymptotes are $\pm\frac{3\sqrt6}{2}$ .

Now, plug in the center of the hyperbola into the point-slope form of the equation of alien to get the equations for the asymptotes.

The center of the hyperbola is (1, 3) .

The equations for the asymptotes are then:

$y-3=\pm\frac{3\sqrt6}{2}(x-1)$

$y=\pm\frac{3\sqrt6}{2}(x-1)+3$

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Example Question #1753 : Pre Calculus

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

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

Possible Answers:

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

$y=\pm\frac{12\sqrt6}{25}(x-5)+3$

$y=\pm\frac{5}{24}(x-5)-3$

$y=\pm\frac{24}{25}(x+5)+3$

Correct answer:

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

Explanation:

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 (h,k) 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 b=5 .

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 (5,-3) .

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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Precalculus : Understand features of hyperbolas and ellipses

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #71 : Understand Features Of Hyperbolas And Ellipses

Example Question #72 : Understand Features Of Hyperbolas And Ellipses

Example Question #1742 : Pre Calculus

Example Question #73 : Understand Features Of Hyperbolas And Ellipses

Example Question #1744 : Pre Calculus

Example Question #1745 : Pre Calculus

Example Question #71 : Hyperbolas And Ellipses

Example Question #71 : Understand Features Of Hyperbolas And Ellipses

Example Question #78 : Understand Features Of Hyperbolas And Ellipses

Example Question #1753 : Pre Calculus

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