A) \[\frac{ab}{2}\]
B) \[ab\]
C) \[2ab\]
D) \[4ab\]
Correct Answer: D
Solution :
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\[\frac{x{{x}_{1}}}{{{a}^{2}}}-\frac{y{{y}_{1}}}{{{b}^{2}}}=2\] ? (i) The equation of the asymptotes are \[\frac{x}{a}-\frac{y}{b}=0\]and\[\frac{x}{a}+\frac{y}{b}=0\] The point of intersection of Eq. (i) with the two asymptotes are given by \[{{X}_{1}}=-\frac{2a}{\frac{{{x}_{1}}}{a}-\frac{{{y}_{1}}}{b}},\,\,{{Y}_{1}}=\frac{2b}{\frac{{{x}_{1}}}{a}-\frac{{{y}_{1}}}{b}}\] and \[{{X}_{2}}=\frac{2a}{\frac{{{x}_{1}}}{a}+\frac{{{y}_{1}}}{b}},\,\,{{Y}_{2}}=\frac{-2b}{\frac{{{x}_{1}}}{a}+\frac{{{y}_{1}}}{b}}\] \[\therefore \]Area of triangle \[=\frac{1}{2}({{X}_{1}}{{Y}_{2}}-{{X}_{2}}{{Y}_{1}})\] \[=\frac{1}{2}\left( \frac{4ab\times 2}{\frac{x_{1}^{2}}{{{a}^{2}}}-\frac{y_{1}^{2}}{{{b}^{2}}}} \right)=4ab\]
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