Question

Find the equation of the hyperbola satisfying the given conditions.
Foci (4, 0), the latus rectum is of length 12.

Answer 1

Here foci are $(\pm 4,0)$ Which lie on x - axis.
So, the equation of hyperbola in stadard form is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\therefore \text{foci}(\pm c,0)is(\pm 4,0)\Rightarrow c=4\text{Length of latus rectum}\frac{2b^2}{a}=12\Rightarrow b^2=6a\text{We know that}{c}^2=a^2+b^2\therefore (4{)}^2=a^2+6a\Rightarrow a^2+6a-16=0\Rightarrow (a+8\left)\right(a-2)=0\Rightarrow a=-8ora=2(\because a=-8\text{is not possible})\Rightarrow a=2\Rightarrow a^2=4\text{Also}{b}^2=6\times 2=12\text{Thus required equation of hyperbola is},\frac{x^2}{4}-\frac{y^2}{12}=1.$