Question

If the straight line $y=mx+c$ touches the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ then prove that $c^2=a^2{m}^2-b^2$

Answer 1

It is given that the line $y=mx+c$ touches the hyperbola.

Consider the tangent of hyperbola in parametric form

$\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$

$\therefore \frac{y\tan \theta }{b}=\frac{x\sec \theta }{a}-1$

$\therefore y=\frac{bx\sec \theta }{a\tan \theta }-\frac{b}{\tan \theta }$

$\therefore y=\frac{b}{a\sin \theta }x-\frac{b}{\tan \theta }$

On comparing with $y=mx+c$, we get

$m=\frac{b}{a\sin \theta }$ and $c=-\frac{b}{\tan \theta }$

$⟹a^2{m}^2-b^2=a^2{\left(\frac{b}{a\sin \theta }\right)}^2-b^2$

$⟹a^2{m}^2-b^2=\frac{b^2}{{\sin }^2\theta }-b^2$

$⟹a^2{m}^2-b^2=\frac{b^2(1-{\sin }^2\theta )}{{\sin }^2\theta }$

$⟹a^2{m}^2-b^2=\frac{b^2{\cos }^2\theta }{{\sin }^2\theta }$

$⟹a^2{m}^2-b^2=\frac{b^2}{{\tan }^2\theta }$

$⟹a^2{m}^2-b^2=c^2$

Hence proved.