GRE Quantitative Reasoning - Hyperbolas

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

$\small \small \small \small \small \small \small \frac{(x-8)^{2}}{4} - \frac{(y-5)^{2}}{12}=1$

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{12} - \frac{(y-8)^{2}}{4}=1$

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{16}=1$

$\small \small \small \small \small \small \small \small \frac{(x-8)^{2}}{12} - \frac{(y-5)^{2}}{16}=1$

Explanation

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small \frac{c}{a}=\frac{2}{1}=\frac{4}{a}$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small a^{2}=4$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small \small 2^{2}+b^{2}=4^{2}$

$\small \small \small b^{2}=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

Find the intersection(s) of the two parabolas: ,

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

$\left(-\frac{1}{2},3\right)$

$\left(\frac{1}{2},3\right)$

$(0,4) \textup{ and }\left(\frac{1}{2},3\right)$

Explanation

Set both parabolas equal to each other and solve for x.

$x=0,-\frac{1}{2}$

Substitute both values of into either parabola and determine .

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

The coordinates of intersection are:

and $\left(-\frac{1}{2},3\right)$

Find the points of intersection:

;

Explanation

To solve, set both equations equal to each other:

To solve as a quadratic, combine like terms by adding/subtracting all three terms from the right side to the left side:

This simplifies to

Solving by factoring or the quadratic formula gives the solutions and .

Plugging each into either original equation gives us:

Our coordinate pairs are and .

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity: $\frac{3}{2}$

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

$\small \small \small \frac{(x-4)^{2}}{16} - \frac{(y-1)^{2}}{20}=1$

$\small \small \small \frac{(x-1)^{2}}{20} - \frac{(y-4)^{2}}{16}=1$

$\small \small \small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{36}=1$

$\small \small \small \small \small \frac{(x-4)^{2}}{20} - \frac{(y-1)^{2}}{36}=1$

Explanation

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small \frac{c}{a}=\frac{3}{2}=\frac{6}{a}$

$\small 12=3a$

$\small a=4$

$\small \small a^{2}=16$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small 4^{2}+b^{2}=6^{2}$

$\small \small b^{2}=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .