Question

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

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

Answer 1

Answer: (B)

Let the parametric equation of the variable line is
$\frac{ x -0}{\cos \theta}=\frac{ y -0}{\sin \theta}= r $
$\Rightarrow x = r \cos \theta ; y = r \sin \theta$
$\therefore $ putting $(x=r \cos \theta ; y=r \sin \theta)$ in $L_{1}=0$, we get
$\frac{1}{ OM }=\frac{(\cos \theta+2 \sin \theta)}{3}\,\,\,\, ...(i)$

$\|$ ly putting the general point in $L_{2}=0$, we get
$\frac{1}{ ON }=\frac{-(\cos \theta+2 \sin \theta)}{4} \,\,\, ...(ii)$
Let $P =( h , k )$ and $OP = r $
$\Rightarrow r \cos \theta= h , r \sin \theta= k$
$\therefore \frac{1}{ OP ^{2}}=\frac{1}{ OM ^{2}}=\frac{1}{ ON ^{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+2 r \sin \theta)^{2}+9( r \cos \theta+2 r \sin \theta)^{2} $
$\Rightarrow 144=16( h +2 k )^{2}+9( h +2 k )^{2}$
$\therefore $ Locus of $P ( h , k )$ is $( x +2 y )^{2}=\frac{144}{25}$