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Precalculus : Understand features of hyperbolas and ellipses

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Example Questions

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 #51 : Graphs

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

Center:(-9,3)

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

Center:(9,-3)

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

Center:(-9,3)

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

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 : 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)

(8, -2)

(2, -8)

(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 #82 : 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:

(5, 7)

(6, 10)

(10, 6)

(14, 50)

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:

(10, 0)(-10, 0)

(0, 12)(0, -12)

(12, 0)(-12, 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 #1753 : 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, 5)(5, -3)

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

(9, 1)(8, 1)

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

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 #83 : Understand Features Of Hyperbolas And Ellipses

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:

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

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

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

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

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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Example Question #1763 : Pre Calculus

Find the center of the hyperbola with the following equation:

25x^2-4y^2+32y-164=0

Possible Answers:

(5, 0)

(0, 4)

(2, 5)

(4, 0)

Correct answer:

(0, 4)

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) .

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

Factor out from the terms.

25x^2-4y^2+32y-164=0

25x^2-4(y^2-8y)-164=0

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

25x^2-4(y^2-8y+16)-164=-64

Add 164 to both sides of the equation.

25x^2-4(y^2-8y+16)=100

Factor the square portion of the equation.

25x^2-4(y-4)^2=100

Divide both sides by 100 to get the standard form of the equation of a hyperbola..

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

For the hyperbola in question, h=0 and k=4 , so the center is at (0, 4) .

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Example Question #88 : Understand Features Of Hyperbolas And Ellipses

Find the center of the hyperbola with the following equation:

4y^2-3x^2-48y-6x+129=0

Possible Answers:

(3, 4)

(6, -1)

(-1, 6)

(4, -3)

Correct answer:

(-1, 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) .

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

Group the terms together and the terms together.

4y^2-3x^2-48y-6x+129=0

4y^2-48y-3x^2-6x+129=0

Factor out from the terms and from the terms.

4(y^2-12y)-3(x^2+2x)+129=0

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

4(y^2-12y+36)-3(x^2+2x+1)+129=141

Subtract 129 from both sides of the equation.

4(y^2-12y+36)-3(x^2+2x+1)=12

Factor the square portions of the equation.

4(y-6)^2-3(x+1)^2=12

Divide both sides by to get the standard form of the equation of a hyperbola.

$\frac{(y-6)^2}{3}-\frac{(x+1)^2}{4}=1$

For the hyperbola in question, h=-1 and k=6 , so the center is at (-1, 6) .

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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 #1752 : Pre Calculus

Example Question #51 : Graphs

Example Question #81 : Hyperbolas And Ellipses

Example Question #82 : Hyperbolas And Ellipses

Example Question #82 : Understand Features Of Hyperbolas And Ellipses

Example Question #1753 : Pre Calculus

Example Question #83 : Understand Features Of Hyperbolas And Ellipses

Example Question #1762 : Pre Calculus

Example Question #1763 : Pre Calculus

Example Question #88 : Understand Features Of Hyperbolas And Ellipses

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