If $e_1$ and $e_2$ are respectively the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola, then the line $\frac{x}{2 e_1}+\frac{y}{2 e_2}=1$ touches the circle having centre at the origin, then its radius is

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Aicharya · Accepted Answer

Hyperbola : $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 ; e_1=\sqrt{\frac{a^2+b^2}{a^2}}$ Conjugate Hyperbola : $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1 ; e_2=\sqrt{\frac{a^2+b^2}{b^2}}$
Line: $\frac{x}{2 e_1}+\frac{y}{2 e_2}=1 \Rightarrow \frac{a^2 x}{2\left(a^2+b^2\right)}+\frac{b^2 y}{2\left(a^2+b^2\right)}=1$ $\Rightarrow a x+b y=2 \sqrt{\left(a^2+b^2\right)}$
Since, the line touches the circle $\Rightarrow$ Distance from origin to line $=$ radius $\Rightarrow \frac{2 \sqrt{\left(a^2+b^2\right)}}{\sqrt{a^2+b^2}}=r \Rightarrow r=2$

If $e_1$ and $e_2$ are respectively the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$…

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