Question

The equation $y=mx+c$ is a tangent to the $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.Then show that $c^2=a^2{m}^2+b^2$.

Answer 1

Substitute $y=mx+C$ into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{a^2}+\frac{(mx+c{)}^2}{b^2}=1$

$(\frac{1}{a^2}+\frac{m^2}{b^2})x^2+\frac{2mc}{b^2}x+\frac{c^2}{b^2}-1=0$

For tangency, $△=0$

${\left(\frac{2mc}{b^2}\right)}^2-4(\frac{1}{a^2}+\frac{m^2}{b^2})(\frac{c^2}{b^2}-1)=0$

$m^2{a}^2{c}^2-(b^2+a^2{m}^2)(c^2-b^2)=0$

$b^2-b^2{c}^2+a^2{b}^2{m}^2=0$

$a^2{m}^2+b^2=c^2$