Pre-Calculus - Understand features of 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?

$\frac{5}{6}$

$\frac{6}{5}$

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

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

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 , 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

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

How can this graph be changed to be the graph of

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

Wrong hyperbola 1

The -intercepts should be at the points and .

The graph should have -intercepts and not -intercepts.

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

The graph should be an ellipse and not a hyperbola.

The -intercepts should be at the points and .

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.

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

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

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

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

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

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

Explanation

The hyperbola pictured is centered at , meaning that the equation has a horizontal shift. The equation must have 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 $\frac{(x-3)^2}{4}$ .

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

and

Explanation

First, we must determine if the major axis is a vertical axis or a horizontal axis. We look at our denominators, and , and see that the larger one is under the -term. Therefore, we know that the greater axis will be a vertical one.

To find out how far the end point are from the center, we simply take $\sqrt{36}=6$ . So we know the end points will be units above and below our center. To find the center, we must remember that for $\frac{(x-h)^2}{w^2}+\frac{(y-k)^2}{v^2}=1$ ,

the center will be .

So for our equation, the center will be . units above and below the center give us and .

Find the endpoints of the major and minor axes of the ellipse described by the following equation:

$\frac{(x-8)^2}{4}+\frac{(y+2)^2}{25}=1$

Explanation

In order to find the endpoints of the major and minor axes of our ellipse, we must first remember what each part of the equation in standard form means:

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

The point given by (h,k) is the center of our ellipse, so we know the center of the ellipse in the problem is (8,-2), and we know that the end points of our major and minor axes will line up with the center either in the x or y direction, depending on the axis. The parts of the equation that will tell us the distance from the center to the endpoints of each axis are and . If we take the square root of each, a will give us the distance from the center to the endpoints in the positive and negative x direction, and b will give us the distance from the center to the endpoints in the positive and negative y direction:

$\frac{(x-8)^2}{4}+\frac{(y+2)^2}{25}=1$

$a^2=4\rightarrow \sqrt{a^2}=\sqrt{4}\rightarrow a=2$

$b^2=25\rightarrow \sqrt{b^2}=\sqrt{25}\rightarrow b=5$

Now it is important to consider the definition of major and minor axes. The major axis of an ellipse is the longer one, will the minor axis is the shorter one. We can see that b=5, which means the axis is longer in the y direction, so this is the major axis. To find the endpoints of the major axis, we'll go 5 units from the center in the positive and negative y direction, respectively, giving us:

Similarly, to find the endpoints of the minor axis, we'll go 2 units from the center in the positive and negative x direction, respectively, giving us:

Find the eccentricity of the hyperbola with the following equation:

$\frac{x^2}{144}-\frac{y^2}{25}=1$

$\frac{13}{12}$

$\frac{12}{13}$

$\frac{144}{25}$

$\frac{25}{144}$

Explanation

Recall the standard form of the equation of a horizontal hyperbola:

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

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{a}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

and

$c=\sqrt{144+25}=\sqrt{169}=13$

Thus,

$e=\frac{13}{12}$

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

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

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

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

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

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

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 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.

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

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

Write the equation for in standard form

$\frac{(x+3)^2 }{4} - \frac{(y-1)^2 } { 9}=1$

$\frac{(x+3)^2 }{9} - \frac{(y+1)^2}{4} = 1$

$\frac{(x+3)^2 }{9} - \frac{(y-1)^2}{4} = 1$

$\frac{(x+3)^2 }{4} - \frac{(y+1)^2 } { 9}=1$

$\frac{(x-3)^2 }{4} - \frac{(y+1)^2}{9} = 1$

Explanation

To write this in standard form, first complete the square for x and y. It will be helpful to re-group our terms:

$9x^2 +54x \enspace \enspace \enspace -4y^2 +8y \enspace \enspace \enspace = -41$ factor out the coefficients on the squared terms

$9(x^2 + 6x \enspace \enspace \enspace ) -4(y^2 -2y \enspace \enspace \enspace) = -41$

Adding 9 will complete the square for x since $(\frac{1}{2}\cdot 6)^2 = 3^2 = 9$

We're adding it in the parentheses, so really we're adding $9\cdot9 = 81$ to both sides

Adding 1 will complete the square for y since $(-2\cdot\frac{1}{2} )^2 = (-1)^2 = 1$

We're really adding $-4\cdot1 = -4$ to both sides

divide both sides by 36

$\frac{(x+3)^2 }{4} - \frac{(y-1)^2 } { 9}=1$

Find the center of the ellipse with the following equation:

$\frac{(x-1)^2}{36}+\frac{(y+2)^2}{9}=1$

Explanation

Recall that the standard form of the equation of an ellipse is

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

For the equation given in the question, and .

The center of the ellipse is at

Express the following equation for a hyperbola in standard form:

$\frac{(y+7)^2}{16}-\frac{(x-13)^2}{9}=1$

$\frac{9(y+7)^2}{16}-\frac{16(x-13)^2}{9}=1$

$\frac{(x+7)^2}{9}-\frac{(y-13)^2}{16}=1$

$\frac{(y+7)^2}{16}-\frac{(x-13)^2}{9}=144$

Explanation

Remember that in order for the equation of a hyperbola to be in standard form, it must be written in one of the following two ways:

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

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

Where the point (h,k) gives the center of the hyperbola. 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. We can see that our equation is of the form given in the second option, as the y term appears first. All we must do to put it in standard form is obtain a 1 on the right side, so we'll divide both sides by 144: