Question

find the equation of the hyperbola satisfying the given condition:

foci (0, ± 12), latus-rectum = 36

Answer 1

The foci of the hyperbola are $$(0,\pm 12)$$ and the latus rectum is 36.
Thus, the value of  $$ae=12$$.

and $$\frac{2b^2}{a}=36$$

$$\Rightarrow b^2=18a$$

Now, using the relation 

$$b^2=a^2(e^2-1)$$,we get:

$$\Rightarrow 18a=144-a^2$$

$$\Rightarrow a^2+18a-144=0$$

$$\Rightarrow (a+24)(a-6)=0$$

$$\Rightarrow a=-24,\text{ or }6$$

$$b^2=-432\text{ or }108$$(but negative value is not possible)\]

Thus, the equation of the hyperbola is

$$\frac{y^2}{36}-\frac{x^2}{108}=1$$.