Question

A variable line having intercepts $e$ and $e^{\prime }$ on co-ordinate axes where $\frac{e}{2},\frac{e^t}{2}$ are eccentricities of a hyperbola and its conjugate hyperbola, then the line touches the circle $x^2+y^2=r^2$, whose radius is

• A. $1$
• B. $2$
• C. $3$
• D. cannot be found

Answer 1

Answer: (B)

$2$

Since $\frac{e}{2},\frac{e^t}{2}$ are eccentricities of hyperbola and its conjugate hyperbola hence, $\frac{4}{e^2}+\frac{4}{{e^{\prime }}^2}=\mathrm{1...}\left(1\right)$ equation of the line is $\frac{x}{e}+\frac{y}{e^{\prime }}=1$

$e^{\prime }x+ey-e{e}^{\prime }=0$ which touches the circle $x^2+y^2=r^2$

Hence, $\frac{e{e}^{\prime }}{\sqrt{e^2+{\left(e^{\prime }\right)}^2}}=r\Rightarrow r^2=4orr=2$ (from (1))