A normal to the hyperbola, 4x2 - 9y2 = 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 :

Question

A normal to the hyperbola, 4x2 - 9y2 = 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 : (A) 4x2 + 9y2 = 121 (B) 9x2 + 4y2 = 169 (C) 4x2 - 9y2 = 121 (D) 9x2 - 4y2 = 169

Tardigrade Admin · Accepted Answer

(3x/sec θ)+(2y/tan θ) = 13 h = (13 sec θ/3) k = (13 tan θ/2) 9x2 - 4y2 = 169

Q. 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 $O A BP$ (O being the origin) is formed, then the locus of P is :

$4 x^{2} + 9 y^{2} = 121$

$9 x^{2} + 4 y^{2} = 169$

$4 x^{2} - 9 y^{2} = 121$

$9 x^{2} - 4 y^{2} = 169$

$\frac{3 x}{sec θ} + \frac{2 y}{t an θ} = 13$
$h = \frac{13 sec θ}{3} k = \frac{13 t an θ}{2}$
$9 x^{2} - 4 y^{2} = 169$

Q. A normal to the hyperbola, 4x2−9y2=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 :

4x2+9y2=121

9x2+4y2=169

4x2−9y2=121

9x2−4y2=169