question_answer

A normal to the hyperbola,\[4{{x}^{2}}-9{{y}^{2}}=36\] meets the co-ordinate axes \[x\] and \[y\] at A and B, respectively. If the parallelogram \[OABP\](O being the origin) is formed, then the locus of \[P\] is          [JEE Online 15-04-2018 (II)]

A)                     \[4{{x}^{2}}-9{{y}^{2}}=121\]     

B) \[4{{x}^{2}}+9{{y}^{2}}=121\]    

C) \[9{{x}^{2}}-4{{y}^{2}}=169\]     

D) \[9{{x}^{2}}+4{{y}^{2}}=169\]

Correct Answer: C

Solution :

Given hyperbola is: \[4{{x}^{2}}-9{{y}^{2}}=36\] Let \[({{x}_{0}},{{y}_{0}})\]be point of contact of normal on the hyperbola Finding slope of normal at that point: Differentiating hyperbola equation we get; \[4.2.x-9.2.y.\frac{dy}{dx}=0\] \[\Rightarrow \frac{dy}{dx}=\frac{4x}{9y}=\]slope of tangent \[\therefore \text{slope of normal=}\frac{-9y}{4x}\] Equation of normal at \[({{x}_{0}},{{y}_{0}})\]is: \[y-{{y}_{0}}=\frac{-9{{y}_{0}}}{4{{x}_{0}}}(x-{{x}_{0}})\] Line intersects X axis at A when \[y=0\] \[\therefore A=\left( \frac{13{{x}_{0}}}{9},0 \right)\] Similarly \[B=\left( 0,\frac{13{{y}_{0}}}{4} \right)\] Given OABP forms a paralleogram\[\to \]diagonals bisect each other (midpoint of diagonals are same) \[\text{midpoint of OB=}\left( \text{0,}\frac{\text{13}{{\text{y}}_{\text{0}}}}{\text{8}} \right)\text{=midpoint of AP}\] \[\text{Let P=(x,y)}\] \[\therefore \]midpoint of \[AP=\left( \frac{\frac{13{{x}_{0}}}{9}+x}{2},\frac{y}{2} \right)\] \[\therefore P(x,y)=\left( \frac{-13{{x}_{0}}}{9},\frac{13{{y}_{0}}}{4} \right)\to 1\] As \[({{x}_{0}},{{y}_{0}})\] lie on hyperbola, it should the equation: \[4{{({{x}_{0}})}^{2}}-9{{({{y}_{0}})}^{2}}=36\] From equation 1: \[{{x}_{0}}=\frac{-9x}{13}\] and \[{{y}_{0}}=\frac{4y}{13}\] Substituting in hyperbola equation, we get: \[9{{x}^{2}}-4{{y}^{2}}=169\] \[\therefore \]locus of point P is hyperbola whose equation is \[:9{{x}^{2}}-4{{y}^{2}}=169\] Hence correct option is \[C.\]