question_answer

The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{5}=1,\] meet x-axis and y-axis at A and B respectively. Then \[{{(OA)}^{2}}-{{(OB)}^{2}},\]where O is the origin, equals:     JEE Main Online Paper (Held On 19 April 2016)

A) \[-\frac{20}{9}\]               

B) \[\frac{19}{9}\]

C) 4                                             

D) \[-\frac{4}{3}\]

Correct Answer: A

Solution :

Given \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{5}=1\] \[\Rightarrow \] \[{{a}^{2}}=4,{{b}^{2}}=5\] \[e=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\frac{4+5}{4}}=\frac{3}{2}\] \[L=\left( 2\times \frac{3}{2},\frac{5}{2} \right)=\left( 3,\frac{5}{2} \right)\] Equation of tangent at \[({{x}_{1}},{{y}_{1}})\]is \[\frac{x{{x}_{1}}}{{{a}^{2}}}-\frac{y{{y}_{1}}}{{{b}^{2}}}=1\] Here \[{{x}_{1}}=3,{{y}_{1}}=\frac{5}{2}\] \[\Rightarrow \] \[\frac{3x}{4}-\frac{y}{2}=1\]\[\Rightarrow \] \[\frac{x}{\frac{4}{3}}+\frac{y}{-2}=1\] x-intercept of the tangent, \[OA=\frac{4}{3}4\] y-intercept of the tangent, OB = ?2 \[O{{A}^{2}}-O{{R}^{2}}=\frac{16}{9}-4=-\frac{20}{9}\]