Q. A normal to the hyperbola, $4x^2 - 9y^2 = 36$ 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 :
Solution:
$\frac{3x}{sec\, \theta}+\frac{2y}{tan\,\theta} = 13$
$h = \frac{13\,sec\,\theta}{3}\quad k = \frac{13\,tan\,\theta}{2}$
$9x^{2} - 4y^{2} = 169$
