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. $2ab$
• D. $4ab$

Answer 1

Answer: (D)

$4ab$

Answer 2

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 tangent from P to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 2$ is $\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}} = 2$ .....(i)

The equation of asymptotes are $\frac{x}{a} - \frac{y}{b}$ = 0 ......(ii)

And $\frac{x}{a} + \frac{y}{b}$ = 0 ....(iii)

The point of intersection of the asymptotes and chord are $\left( \frac{2a}{x_{1}/a - y_{1}/b},\frac{2b}{x_{1}/a - y_{1}/b} \right);\left( \frac{2a}{x_{1}/a + y_{1}/b},\frac{- 2b}{x_{1}/a + y_{1}/b} \right)$, (0, 0)

$\therefore$ Area of triangle = $\frac{1}{2}|(x_{1}y_{2} - x_{2}y_{1})|$ =

$\frac{1}{2}\left| \left( \frac{- 8ab}{x_{1}^{2}/a^{2} - y_{1}^{2}/b^{2}} \right) \right| = 4ab$