Question

Find the equation of the hyperbola satisfying the given conditions.
Foci $(\pm 3\sqrt{5},0)$ the latus rectum is of length 8.

Answer 1

Here foci are $(\pm 3\sqrt{5},0)$ which lie on x - axis
So the equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\therefore \text{foci}(\pm c,0)is(\pm 3\sqrt{5},0)\Rightarrow c=3\sqrt{5}\text{Length of latus rectum}\frac{2b^2}{a}=8\Rightarrow b^2=4a\text{We know that}{c}^2=a^2+b^2\therefore (3\sqrt{5}{)}^2=a^2+4a\Rightarrow a^2+4a-45=0\Rightarrow (a+9\left)\right(a-5)=0\Rightarrow a=-9ora=5\because a=-9\text{is not possible}$