From any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, tangents are drawn to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=2$. The area of the figure formed by the chord of contact of that point and the asymptotes is

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Aicharya · Accepted Answer

Let tangents are drawn from $A(a, 0)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=2$ $P Q$ is the chord of contact. Asymptotes are $A_1$ and $A_2$. Equation of $A_1$ and $A_2$ are $\frac{x}{a}-\frac{y}{b}=0$ and $\frac{x}{a}+\frac{y}{b}=0$ respectively. Equation of $P Q$ is given by $\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=2$ $\Rightarrow \frac{a x}{a^2}-0=2 \Rightarrow x=2 a$ Now solving $x=2 a$ and asymptotes $A_1$ and $A_2$ we get $P \equiv(2 a, 2 b)$ and $Q \equiv(2 a,-2 b)$ $\begin{aligned} & \Rightarrow \quad M \text { is the mid-point of } P Q \\ & \Rightarrow \quad M \equiv(2 a, 0) \\ & \Rightarrow \quad O M=2 a \text { and } P Q=4 b\end{aligned}$ $\therefore$ Area $(\triangle O P Q)=\frac{1}{2} \times$ base $\times$ height $\begin{aligned} & =\frac{1}{2} \times P Q \times O M=\frac{1}{2} \times 4 b \times 2 a \\ & =4 a b\end{aligned}$

From any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, tangents are drawn to the hyperbola…

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