Question

Let $P$ and $Q$ be any two points on the lines represented by $2x-3y=0$ and $2x+3y=0$ 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. $4x^2-9y^2+30=0$
• B. $9x^2-4y^2+30=0$
• C. $9x^2-4y^2-30=0$
• D. $4x^2-9y^2-30=0$

Answer 1

Answer: (A), (D)

We have
$\mathrm{Area}(\Delta OPQ)=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ a& \frac{2a}{3}& 1\\ b& \frac{-2b}{3}& 1\end{array}\right|=5$
(Given)

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