Question

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 cut-off by the chord of contact on the asymptotes is equal to

• A. $\frac{ab}{2}$
• B. ab
• C. 2 ab
• D. 4 ab

Answer 1

Answer: (D)

4 ab

Let $P(x_1,y_1)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Then,
$\frac{{x_1}^2}{a^2}-\frac{{y_1}^2}{b^2}=1$
The chord of contact of tangents from $P$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=2$ is
$\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=2\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(i\right)$

The equations of the asymptotes are
$\frac{x}{a}-\frac{y}{b}=0$ and $\frac{x}{a}+\frac{y}{b}=0$

The points of intersection of $\left(i\right)$ with the two asymptotes are given by
$x_1=\frac{2a}{\frac{x_1}{a}-\frac{y_1}{b}},y_1=\frac{2b}{\frac{x_1}{a}-\frac{y_1}{b}}$
$x_2=\frac{2a}{\frac{x_1}{a}-\frac{y_1}{b}},y_2=\frac{-2b}{\frac{x_1}{a}-\frac{y_1}{b}}$
$\therefore$ Area of the triangle
$\frac{1}{2}(x_1{y}_2-x_2{y}_1)=\frac{1}{2}\left(\frac{4ab\times 2}{\frac{{x_1}^2}{a^2}+\frac{{y_1}^2}{b^2}}\right)=4ab$