Question

If $ax+by=1$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ then the value of $a^2–b^2$ is

• A. $b^2{e}^2$
• B. $\frac{1}{b^2{e}^2}$
• C. $a^2{e}^2$
• D. $\frac{1}{a^2{e}^2}$

Answer 1

The correct option is D $\frac{1}{a^2{e}^2}$

Equation of tangent to the hyperbola,
$y=mx+\sqrt{a^2{m}^2-b^2}$
Given equation of the tangent is,
$ax+by=1\Rightarrow y=-\frac{a}{b}x+\frac{1}{b}$
Comparing both the equation,
$m=-\frac{a}{b}\sqrt{a^2{m}^2-b^2}=\frac{1}{b}\Rightarrow a^2\times \frac{a^2}{b^2}-b^2=\frac{1}{b^2}\Rightarrow (a^2-b^2)(a^2+b^2)=1$
Eccentricity of hyperbola is,
$e^2=1+\frac{b^2}{a^2}\Rightarrow a^2{e}^2=a^2+b^2$
So,
$a^2-b^2=\frac{1}{a^2{e}^2}$