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Precalculus : Conic Sections

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Example Questions

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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Example Question #1752 : Pre Calculus

Find the equations for the asymptotes of the following hyperbola:

$\frac{(x-5)^2}{16}-\frac{(y+3)^2}{4}=1$

Possible Answers:

$y=\frac{1}{4}x-\frac{17}{4}$ AND $y=-\frac{1}{4}x-\frac{17}{4}$

$y=\frac{1}{4}x-\frac{17}{4}$

$y=\frac{1}{2}(x)-\frac{11}{2}$

$y=-\frac{1}{2}(x)-\frac{11}{2}$

$y=\frac{1}{2}(x)-\frac{11}{2}$ AND $y=-\frac{1}{2}(x)-\frac{11}{2}$

Correct answer:

$y=\frac{1}{2}(x)-\frac{11}{2}$ AND $y=-\frac{1}{2}(x)-\frac{11}{2}$

Explanation:

The standard form of a hyperbola is given by

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The equations for the asymptotes of a hyperbola are given by

$y=k\pm\frac{b}{a}(x-h)$

Since our equation is already in standard form, we know h=5, k=-3, and

$a^2=16\Rightarrow a=4$

$b^2=4\Rightarrow b=2$

Plugging these vaules into the equation for the asymptote gives

$y=-3\pm \frac{2}{4}(x-5)$

$y=-3\pm \frac{1}{2}(x-5)$

$y=-3\pm \frac{1}{2}(x)-\frac{1}{2}(5)$

$y=-3\pm \frac{1}{2}(x)-\frac{5}{2}$

$y=\pm \frac{1}{2}(x)-\frac{6}{2}-\frac{5}{2}$

$y=\pm \frac{1}{2}(x)-\frac{11}{2}$

So, the equations for the asymptotes are given by

$y=\frac{1}{2}(x)-\frac{11}{2}$ AND $y=-\frac{1}{2}(x)-\frac{11}{2}$

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Example Question #1 : Hyperbolas

Find the center and the vertices of the following hyperbola:

$\frac{(x+9)^2}{25}-\frac{(y-3)^2}{16}=1$

Possible Answers:

Center:(9,-3)

Vertices:(-13,3),(-5,3)

Center:(-9,3)

Vertices:(-14,3),(-4,3)

Center:(-9,3)

Vertices:(-9,7),(-9,-1)

Center:(-9,3)

Vertices:(-14,7),(-4,7)

Center:(9,-3)

Vertices:(-9,8),(-9,-2)

Correct answer:

Center:(-9,3)

Vertices:(-14,3),(-4,3)

Explanation:

In order to find the center and the vertices of the hyperbola given in the problem, we must examine the standard form of a hyperbola:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

The point (h,k) gives the center of the hyperbola. We can see that the equation in this problem resembles the second option for standard form above, so right away we can see the center is at:

Center:(-9,3)

In the first option, where the x term is in front of the y term, the hyperbola opens left and right. In the second option, where the y term is in front of the x term, the hyperbola opens up and down. In either case, the distance tells how far above and below or to the left and right of the center the vertices of the hyperbola are. Our equation is in the first form, where the x term is first, so the hyperbola opens left and right, which means the vertices are a distance to the left and right of the center. We can now calculate by identifying it in our equation, and then go 3 units to the left and right of our center to find the following vertices:

$\frac{(x+9)^2}{25}-\frac{(y-3)^2}{16}=1$

$a^2=25\rightarrow a=5$

Vertices:(-14,3),(-4,3)

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Example Question #81 : Understand Features Of Hyperbolas And Ellipses

Find the center of the hyperbola with the following equation:

$\frac{(x-2)^2}{9}-\frac{(y+8)^2}{16}=1$

Possible Answers:

(2, -8)

(-2, 8)

(8, -2)

(3, 4)

Correct answer:

(2, -8)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola in question, h=2 and k=-8 , so the center is at (2, -8) .

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Example Question #81 : Understand Features Of Hyperbolas And Ellipses

Find the center of the hyperbola with the following equation:

$\frac{(y-6)^2}{14}-\frac{(x-10)^2}{50}=1$

Possible Answers:

(14, 50)

(6, 10)

(10, 6)

(5, 7)

Correct answer:

(10, 6)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola in question, h=10 and k=6 , so the center is at (10, 6) .

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Example Question #82 : Understand Features Of Hyperbolas And Ellipses

Find the vertices of the hyperbola with the following equation:

$\frac{x^2}{144}-\frac{y^2}{100}=1$

Possible Answers:

(12, 0)(-12, 0)

(0, 12)(0, -12)

(10, 0)(-10, 0)

(0, 10)(0, -10)

Correct answer:

(12, 0)(-12, 0)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the vertices are located at (h-a, k)(h+a, k) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the vertices are located at (h, k+b)(h, k-b) .

For the hyperbola in question, the center is located at (0,0) and a=12 . The vertices must be at (12, 0) and (-12, 0) .

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Example Question #1757 : Pre Calculus

Find the vertices of a hyperbola with the following equation:

$\frac{(x-5)^2}{5}-\frac{(y-1)^2}{16}=1$

Possible Answers:

$(5+\sqrt5, 1)(5-\sqrt5, 1)$

(5, 5)(5, -3)

(9, 1)(8, 1)

$(5, 1+\sqrt5)(5, 1-\sqrt5)$

Correct answer:

$(5+\sqrt5, 1)(5-\sqrt5, 1)$

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the vertices are located at (h-a, k)(h+a, k) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the vertices are located at (h, k+b)(h, k-b) .

For the hyperbola in question, the center is located at (5, 1) and $a=\sqrt5$ . The vertices must be at $(5+\sqrt5, 1)$ and $(5-\sqrt5, 1)$ .

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Example Question #1761 : Pre Calculus

Find the endpoints of the conjugate vertices of the hyperbola with the following equation:

$\frac{(y-9)^2}{12}-\frac{(x+1)^2}{3}=1$

Possible Answers:

$(-1+\sqrt3, 9)(-1-\sqrt3, 9)$

$(2\sqrt3, \sqrt3)(-2\sqrt3, \sqrt3)$

$(1, 9+2\sqrt3)(1, 9-2\sqrt3)$

$(-1+2\sqrt3, 9)(-1-2\sqrt3, 9)$

Correct answer:

$(-1+\sqrt3, 9)(-1-\sqrt3, 9)$

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the endpoints of the conjugate axis are located at (h, k+b)(h, k-b) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the endpoints of the conjugate axis are located at (h-a, k)(h+a, k) .

For the hyperbola in question, the center is located at (-1, 9) and $a=\sqrt3$ . The endpoints of the conjugate axis must be at $(-1+\sqrt3, 9)$ and $(-1-\sqrt3, 9)$ .

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Example Question #1762 : Pre Calculus

Find the endpoints of the conjugate axis for the hyperbola with the following equation:

$\frac{(x-5)^2}{9}-\frac{y^2}{25}=1$

Possible Answers:

(5, 5)(5, -5)

(5, 0)(10, 5)

(-5, -5)(5, 5)

(-5,0)(-10, 0)

Correct answer:

(5, 5)(5, -5)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the endpoints of the conjugate axis are located at (h, k+b)(h, k-b) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the endpoints of the conjugate axis are located at (h-a, k)(h+a, k) .

For the hyperbola in question, the center is located at (5, 0) and b=5 . The endpoints of the conjugate axis must be at (5, 5) and (5, -5) .

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Precalculus : Conic Sections

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #78 : Understand Features Of Hyperbolas And Ellipses

Example Question #1753 : Pre Calculus

Example Question #1752 : Pre Calculus

Example Question #1 : Hyperbolas

Example Question #81 : Understand Features Of Hyperbolas And Ellipses

Example Question #81 : Understand Features Of Hyperbolas And Ellipses

Example Question #82 : Understand Features Of Hyperbolas And Ellipses

Example Question #1757 : Pre Calculus

Example Question #1761 : Pre Calculus

Example Question #1762 : Pre Calculus

All Precalculus Resources

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