Question

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

• A. $4x^2-9y^2=121$
• B. $4x^2+9y^2=121$
• C. $9x^2-4y^2=169$
• D. $9x^2+4y^2=169$

Answer 1

Answer: (C)

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

given hyperbola is: $4x^2-9y^2=36$

let $(x_0,y_0)$ be point of contact of normal on the hyperbola

Finding slope of normal at that point:

Differntiating hyperbola equation we get; $4\times 2\times x-9\times 2\times y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{4x}{9y}$ = slope of tangent

$\therefore$ slope of normal=$\frac{-9y}{4x}$

equation of normal at $(x_0,y_0)$ is:

$y-y_0=\frac{-9y_0}{4x_0}(x-x_0)$

line intersects X axis at A when y=0

$\therefore A=(\frac{13x_0}{9},0)$

similarly $B=(0,\frac{13y_0}{4})$

given OABP forms a paralleogram$\to$diagonals bisect each other(midpoint of diagonals are same)

midpoint of OB=$(0,\frac{13y_0}{8})$=midpoint of AP

Let $P=(x,y)$

$\therefore midpointofAP=(\frac{\frac{13x_0}{9}+x}{2},\frac{y}{2})$

$\therefore P(x,y)=(\frac{-13x_0}{9},\frac{13y_0}{4})\to 1$

As $(x_0,y_0)$ lie on hyperbola,it shoud the its equation:

$4{\left(x_0\right)}^2-9{\left(y_0\right)}^2=36$

from equation 1: $x_0=\frac{-9x}{13}$ and $y_0=\frac{4y}{13}$

substituting in hyperbola equation ,we get:

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

$\therefore$ locus of point P is hyperbola whose equation is:$9x^2-4y^2=169$

hence correct option is C.