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A normal to the hyperbola, \mathrm{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

Option: 1

\mathrm{9 x^2+4 y^2=169}


Option: 2

\mathrm{4 x^2-9 y^2=121}


Option: 3

\mathrm{4 x^2+9 y^2=121}


Option: 4

\mathrm{9 x^2-4 y^2=169}


Answers (1)

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Given hyperbola is \mathrm{4 x^2-9 y^2=36}

\mathrm{ \Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=1 }                                      ...(i)

Equation of normal to (i) is given by

\mathrm{ \frac{3 x}{\sec \theta}+\frac{2 y}{\tan \theta}=9+4 \Rightarrow \frac{3 x}{\sec \theta}+\frac{2 y}{\tan \theta}=13 }

\therefore Coordinates of A are \mathrm{\left(\frac{13}{3} \sec \theta, 0\right)} and coordinates of

B are \mathrm{\left(0, \frac{13}{2} \tan \theta\right)}

Let the coordinates of P be (h, k). Since, diagonals of parallelogram bisect each other.

\mathrm{ \begin{aligned} & \therefore\left(\frac{h+\frac{13}{3} \sec \theta}{2}, \frac{0+k}{2}\right)=\left(0, \frac{13}{4} \tan \theta\right) \\ & \Rightarrow h=-\frac{13}{3} \sec \theta, k=\frac{13}{2} \tan \theta \end{aligned} }

\mathrm{\Rightarrow \sec \theta=-\frac{3 h}{13} \quad \ldots (ii)\, \, and\, \, \tan \theta=\frac{2 k}{13}}                   ...(iii)

Squaring (ii) and (iii) and then subtracting, we get

\mathrm{ \frac{9 h^2}{169}-\frac{4 k^2}{169}=1 \Rightarrow 9 h^2-4 k^2=169 }

Thus, locus of P is \mathrm{9 x^2-4 y^2=169.}

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