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

Option: 1

\mathrm{\frac{ab}{2}}


Option: 2

\mathrm{ab}


Option: 3

\mathrm{2\ ab}


Option: 4

\mathrm{4\ ab}


Answers (1)

best_answer

Let \mathrm{P(x_{1},y_{1})} be a point on the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text {, then } \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1}
The chord of contact of tangent from \mathrm{P} to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=2 \quad \text { is } \quad \frac{x x_1}{a^2}-\frac{y y_1}{b^2}=2\ \ ..........(i)}
The equation of asymptotes are \mathrm{\frac{x}{a}-\frac{y}{b}=0\ \ ..........(ii)}
\mathrm{\text { And } \frac{x}{a}+\frac{y}{b}=0\ \ ..........(iii)}
The point of intersection of the asymptotes and chord are
\mathrm{\left(\frac{2 a}{x_1 / a-y_1 / b}, \frac{2 b}{x_1 / a-y_1 / b}\right) ;\left(\frac{2 a}{x_1 / a+y_1 / b}, \frac{-2 b}{x_1 / a+y_1 / b}\right),(0,0)}
\therefore Area of triangle \mathrm{=\frac{1}{2}\left|\left(x_1 y_2-x_2 y_1\right)\right|=\frac{1}{2}\left|\left(\frac{-8 a b}{x_1^2 / a^2-y_1^2 / b^2}\right)\right|=4 a b}

 

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Divya Prakash Singh

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