how to find hyperbola equation knowing tangent line and point

Question

I have a problem.

A hyperbola passes through point $(3,2)$ and $9x+2y-15=0$ is a tangent line. Find the equation of hyperbola.

Answer 1

Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be the hyperbola.It passes through $(3,2)$.

So $\frac{9}{a^2}-\frac{4}{b^2}=1................(1)$

Also tangent $9x+2y-15=0$ is given.

Put $y=\frac{-9x+15}{2}$ in the equation of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{x^2}{a^2}-\frac{1}{b^2}\times\frac{81x^2+225-270x}{4}=1$

$(\frac{1}{a^2}-\frac{81}{4b^2})x^2+\frac{270}{4b^2}x-(1+\frac{225}{4b^2})=0$

This is a quadratic equation in $x$,whose discriminant should be zero,as tangent cuts hyperbola at one point.Put $b^2-4ac=0$

$\frac{18225}{4b^4}=-4(\frac{1}{a^2}-\frac{81}{4b^2})(1+\frac{225}{4b^2}).....(2)$

Solve $(1)$ and $(2)$ to find $a^2$ and $b^2$,and substitute in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to get the equation.

Hope it helps.