Question

Let $P$ and $Q$ be any two points on the lines represented by $2 x-3 y=0$ and $2 x+3 y=0$ respectively. If the area of triangle OPQ (where $O$ is origin) is 5 , then the equation of the locus of mid-point of $PQ$, can be equal to

• A. $ 4 x ^2-9 y ^2+30=0$
• B. $9 x ^2-4 y ^2+30=0$
• C. $9 x^2-4 y^2-30=0$
• D. $4 x^2-9 y^2-30=0$

Answer 1

Answer: (A), (D)

We have
$\operatorname{Area}(\Delta OPQ )=\frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ a & \frac{2 a }{3} & 1 \\ b & \frac{-2 b }{3} & 1\end{vmatrix}=5$
(Given)

$\Rightarrow \frac{4 a b}{3}= \pm 10$
So, $ 4 ab = \pm 30$
Also $2 h = a + b$
$2 k =\frac{2 a -2 b }{3} \Rightarrow a - b =3 k$
As $ 4 a b=(a+b)^2-(a-b)^2$
$\Rightarrow \pm 30=4 h ^2-9 k ^2$ [Using (1), (2) and (3)]
So, required locus can be $4 x^2-9 y^2= \pm 30$, which is hyperbola