Question

The line $2x+y=4$ touches the hyperbola whose foci are $(\pm 3,0)$. Find the equation of the hyperbola.

Answer 1

Line $2x+y=4$ foci $(\pm 3,0)$

$\frac{2x+y}{4}=\frac{4}{4}\Rightarrow \frac{x}{2}+\frac{y}{4}=1a=2$

foci $(\pm 3,0)i.e.(\pm ae,0)\therefore ae=3e=\frac{3}{2}$

Let $e$ be the eccentricity of the required hyperbola

$\frac{x^2}{{a^{\prime }}^2}-\frac{y^2}{{b^{\prime }}^2}=1\therefore$ coordinates are

$a^{\prime }{e}^{\prime }=\frac{3}{2}\Rightarrow \frac{3}{2}{a}^{\prime }=3\Rightarrow a^{\prime }=2$

$b^{\prime 2}=a^{\prime 2}(e^{\prime 2}-1)=4[\frac{9}{4}-1]=4\left[\frac{5}{4}\right]=5\Rightarrow \frac{x^2}{4}-\frac{y^2}{25}=1$