Question

Which is the equation of a hyperbola with directrices at y = ±2 and foci at (0, 6) and (0, −6)?

Answer 1

Given:

Hyperbola with directrices at y = ±2 and foci at (0, 6) and (0, −6).

To find:

The equation of hyperbola.

Solution:

We have, directrices at y = ±2 so this hyparabola is along the y-axis.

The standard form of hyperbola is

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

where, (h,k) is center, foci are $(h,k\pm c)$ and directrix are $y=k\pm \dfrac{a^2}{c}$.

On comparing foci, we get

$(h,k\pm c)=(0,\pm 6)$

$h=0,k=0,c=6$

On comparing directrix we get

$k\pm \dfrac{a^2}{c}=\pm 2$

$\dfrac{a^2}{c}=2$

$\dfrac{a^2}{6}=2$

$a^2=12$

Now,

$a^2+b^2=c^2$

$12+b^2=(6)^2$

$b^2=36-12$

$b^2=24$

Putting $h=0,k=0,a^2=12,b^2=24$, we get

$\dfrac{(y-0)^2}{12}-\dfrac{(x-0)^2}{24}=1$

$\dfrac{y^2}{12}-\dfrac{x^2}{24}=1$

Therefore, the equation of hyperbola is $\dfrac{y^2}{12}-\dfrac{x^2}{24}=1$.