Question

A hyperbola has its foci at (1, 7) and (1, –13). a directrix of the hyperbola is y = . what is the equation of the hyperbola?

Answer 1

The equation of the hyperbola is $\frac{(y + 3)^2}{64} -\frac{(x - 1)^2}{36} = 1$

How to determine the hyperbola equation?

The foci are given as:

(1, 7) and (1, -13)

The foci of a hyperbola are represented using: (m, n+p) and (m, n-p)

Where the center of the hyperbola is (n,p)

By comparison, we have:

m = 1

n + p = 7

n - p =13

Where n = ae

The solution to the above system of equations is

m = 1, n = 10 and p = -3

The directrix is given as:

y = 64/10

The directrix is represented as:

y = a/e

So, we have:

a/e = 64/10

Substitute n = ae in n = 10

ae = 10

Solving ae = 10 and a/e = 64/10, we have:

a = 8 and e = 10/8

To calculate b, we make use of:

$b = a\sqrt{(e^2 - 1)$

So, we have:

$b = 8\sqrt{(\frac{10}{8}^2 - 1)$

Evaluate

b = 8 * 0.75

b = 6

The equation of the hyperbola is then calculated using:

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

This gives

$\frac{(y + 3)^2}{8^2} -\frac{(x - 1)^2}{6^2} = 1$

Evaluate

$\frac{(y + 3)^2}{64} -\frac{(x - 1)^2}{36} = 1$

Hence, the equation of the hyperbola is $\frac{(y + 3)^2}{64} -\frac{(x - 1)^2}{36} = 1$

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