If e_1 and e_2 are respectively the eccentric | vedclass

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(A) For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,the eccentricity is $e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \frac{\sqrt{a^2+b^2}}{a}$.For its co

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$2$

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(A) For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,the eccentricity is $e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \frac{\sqrt{a^2+b^2}}{a}$.For its conjugate hyperbola $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$,the eccentricity is $e_2 = \sqrt{1 + \frac{a^2}{b^2}} = \frac{\sqrt{a^2+b^2}}{b}$.The given line is $\frac{x}{2 e_1} + \frac{y}{2 e_2} = 1$.Substituting the values of $e_1$ and $e_2$:$\frac{x}{2(\frac{\sqrt{a^2+b^2}}{a})} + \frac{y}{2(\frac{\sqrt{a^2+b^2}}{b})} = 1$$\frac{ax}{2\sqrt{a^2+b^2}} + \frac{by}{2\sqrt{a^2+b^2}} = 1$$ax + by = 2\sqrt{a^2+b^2}$.Since this line touches the circle $x^2 + y^2 = r^2$ with center at the origin,the perpendicular distance from the origin $(0,0)$ to the line must equal the radius $r$.$r = \frac{|0 + 0 - 2\sqrt{a^2+b^2}|}{\sqrt{a^2+b^2}} = \frac{2\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} = 2$.Thus,the radius is $2$.

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

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