Question

A variable line $L$ is drawn through $O(0,0)$ to meet the lines $L_1:x+2y-3=0$ and $L_2:x+2y+4=0$ at points $M$ and $N$ respectively. A point $P$ is taken on line $L$ such that $\frac{1}{O{P}^2}=\frac{1}{O{M}^2}+\frac{1}{O{N}^2}$. Then the locus of $P$ is

• A. $x^2+4y^2=\frac{144}{25}$
• B. $(x+2y{)}^2=\frac{144}{25}$
• C. $4x^2+y^2=\frac{144}{25}$
• D. $(x-2y{)}^2=\frac{144}{25}$

Answer 1

The correct option is B $(x+2y{)}^2=\frac{144}{25}$

Let the parametric equation of the variable line $L$ be $\frac{x-0}{\cos \theta }=\frac{y-0}{\sin \theta }=r$, where $OP=r$
Let the coordinates of $P$ be $(h,k)\equiv (r\cos \theta ,r\sin \theta )$


Putting $x=r\cos \theta ;y=r\sin \theta$ in $L_1$, we get
$\frac{1}{OM}=\frac{\cos \theta +2\sin \theta }{3}$
Similarly, putting the general point in $L_2$, we get
$\frac{1}{ON}=-\left(\frac{\cos \theta +2\sin \theta }{4}\right)$

Given, $\frac{1}{O{P}^2}=\frac{1}{O{M}^2}+\frac{1}{O{N}^2}$
$\Rightarrow \frac{1}{r^2}=\frac{(\cos \theta +2\sin \theta {)}^2}{9}+\frac{(\cos \theta +2\sin \theta {)}^2}{16}$
$\Rightarrow 144=16(r\cos \theta +2r\sin \theta {)}^2+9(r\cos \theta +2r\sin \theta {)}^2$
$\Rightarrow 144=16(h+2k{)}^2+9(h+2k{)}^2$
$\therefore$ Locus of $P(h,k)$ is $(x+2y{)}^2=\frac{144}{25}$