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Precalculus : Hyperbolas and Ellipses

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Example Questions

Example Question #121 : Hyperbolas And Ellipses

Determine the length of the foci for the following hyperbola equation:

$\frac{(x-1)^2}{9}+\frac{(x-4)^2}{16}=1$

Possible Answers:

$\sqrt{5}$

Correct answer:

Explanation:

$\frac{(x-1)^2}{9}+\frac{(x-4)^2}{16}=1$

To solve, simply use the follow equation where c is the length of the foci.

In this particular case,

$a=\sqrt{9}=3$

$b=\sqrt{16}=4$

Thus,

$c^2=a^2+b^2\Rightarrow c=\sqrt{a^2+b^2}$

$c=\sqrt{3^2+4^2}=5$

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Example Question #122 : Hyperbolas And Ellipses

Find the foci of the hyperbola with the following equation:

$\frac{(x-5)^2}{16}-\frac{(y+3)^2}{33}=1$

Possible Answers:

(12, 3) and (2, 3)

(12, 4) and (-2, -10)

(5, 4) and (5, -10)

(17, -3) and (-17, -3)

(12, -3) and (-2, -3)

Correct answer:

(12, -3) and (-2, -3)

Explanation:

The standard form of the equation for a hyperbola is given by

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

The foci are located at (h+c, k) and (h-c, k), where c is found by using the formula

c^2=a^2+b^2

Since our equation is already in standard form, you can see that

a^2=16 , b^2=33

Plugging into the formula

$c^2=49 \Rightarrow c=7$

So the foci are found at

$(5+7,-3) \Rightarrow (12,-3)$ AND

$(5-7,-3) \Rightarrow (-2,-3)$

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Example Question #1791 : Pre Calculus

How can this graph be changed to be the graph of

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

Wrong hyperbola 1

Possible Answers:

The -intercepts should be at the points (3,0) and (-3,0) .

The center box should extend up to and down to , stretching the graph.

The graph should have -intercepts and not -intercepts.

The graph should be an ellipse and not a hyperbola.

The -intercepts should be at the points (1,0) and (-1,0) .

Correct answer:

The -intercepts should be at the points (1,0) and (-1,0) .

Explanation:

This equation should be thought of as $\frac{x^2}{1^2}-\frac{y^2}{3^2}=1$ .

This means that the hyperbola will be determined by a box with x-intercepts at $\pm 1$ and y-intercepts at $\pm3$ .

The hyperbola was incorrectly drawn with the intercepts at $\pm2$ instead.

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Example Question #121 : Hyperbolas And Ellipses

Which of the following would NOT be true of the graph for $\frac{(y-2)^2}{49}-\frac{(x+1)^2}{9}=1$ ?

Possible Answers:

The graph is centered at (-1,2) .

The graph never intersects with the -axis.

The graph opens up and down.

All of these statements are true.

The graph never intersects with the -axis.

Correct answer:

The graph never intersects with the -axis.

Explanation:

The graph should look like this:

Right hyperbola 2

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Example Question #124 : Hyperbolas And Ellipses

Which of these equations produce this graph, rotated 90 degrees?

Wrong hyperbola 1

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Rotated 90 degrees, this graph would be opening up and down instead of left and right, so the equation will have the y term minus the x term.

The box that the hyperbola is drawn around will also rotate. It will now be up/down 2 and left/right 3.

This makes the correct equation

$\frac{y^2}{4}-\frac{x^2}{9}=1$ .

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Example Question #125 : Hyperbolas And Ellipses

What is the equation of the conic section graphed below? Right hyperbola 1

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The hyperbola pictured is centered at (3,0) , meaning that the equation has a horizontal shift. The equation must have (x-3) rather than just x. The hyperbola opens up and down, so the equation must be the y term minus the x term. The hyperbola is drawn according to the box going up/down 5 and left/right 2, so the y term must be $\frac{y^2}{5^2}$ or $\frac{y^2}{25}$ , and the x term must be $\frac{(x-3)^2}{2^2}$ or $\frac{(x-3)^2}{4}$ .

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Example Question #122 : Hyperbolas And Ellipses

The equation of an ellipse, , is $\frac{(x+3)^2}{9}+\frac{(y-7)^2}{36}=1$ . Which of the following is the correct eccentricity of this ellipse?

Possible Answers:

$\frac{6}{5}$

$\frac{2}{\sqrt{3}}$

$\frac{\sqrt{3}}{2}$

$\frac{5}{6}$

Correct answer:

$\frac{\sqrt{3}}{2}$

Explanation:

The equation for the eccentricity of an ellipse is $e=\frac{c}{a}$ , where is eccentricity, is the distance from the foci to the center, and is the square root of the larger of our two denominators.

Our denominators are and , so $a=\sqrt{36}=6$ .

To find , we must use the equation a^2-b^2=c^2 , where is the square root of the smaller of our two denominators.

This gives us $36-9=27$ , so $c=\sqrt{27}=\sqrt{9}\cdot \sqrt{3}=3\sqrt{3}$ .

Therefore, we can see that

$e=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}$ .

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Precalculus : Hyperbolas and Ellipses

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #121 : Hyperbolas And Ellipses

Example Question #122 : Hyperbolas And Ellipses

Example Question #1791 : Pre Calculus

Example Question #121 : Hyperbolas And Ellipses

Example Question #124 : Hyperbolas And Ellipses

Example Question #125 : Hyperbolas And Ellipses

Example Question #122 : Hyperbolas And Ellipses

All Precalculus Resources

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