Solveeit Logo
Login

Question

Question: From any point on the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1,tangents are d...

From any point on the hyperbola x2a2\frac{x^{2}}{a^{2}} – y2b2\frac{y^{2}}{b^{2}} = 1,tangents are drawn to the hyperbola x2a2\frac{x^{2}}{a^{2}} – y2b2\frac{y^{2}}{b^{2}} = 2. The area cut off by the chord of contact on the asympotes is equal to –

A

ab/2

B

ab

C

2ab

D

4ab

Answer

4ab

Explanation

Solution

Let P(x1, y1) be a point on the hyperbola x2a2−y2b2\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, then x12a2−y12b2\frac{{x_{1}}^{2}}{a^{2}} - \frac{{y_{1}}^{2}}{b^{2}} = 1

The chord of contact of tangents from P to the hyperbola x2a2−y2b2\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 2 is xx1a2−yy1b2\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}}= 2 …(i)

The equation of the asympotes are xa−yb\frac{x}{a} - \frac{y}{b} = 0 and xa+yb\frac{x}{a} + \frac{y}{b} = 0

The point of intersection of (i) with the two asymptotes are given by

x¢ = 2ax1a−y1b\frac{2a}{\frac{x_{1}}{a} - \frac{y_{1}}{b}}, y¢ = 2bx1a−y1b\frac{2b}{\frac{x_{1}}{a} - \frac{y_{1}}{b}}, x¢¢ = 2ax1a+y1b\frac{2a}{\frac{x_{1}}{a} + \frac{y_{1}}{b}},

y¢¢ = −2ax1a+y1b\frac{- 2a}{\frac{x_{1}}{a} + \frac{y_{1}}{b}}

\ Area of the triangle = 12\frac{1}{2} (x1y2 – x2y1)

= 12\frac { 1 } { 2 } [4ab×2x12a2−y12b2]\left\lbrack \frac{4ab \times 2}{\frac{x_{1}^{2}}{a^{2}} - \frac{y_{1}^{2}}{b^{2}}} \right\rbrack = 4ab.