Question

A variable line $L=0$ 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 $L=0$ such that $\frac{1}{O{P}^2}=\frac{1}{O{M}^2}+\frac{1}{O{N}^2}$. 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

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
$\frac{1}{OM}=\frac{(\cos \theta +2\sin \theta )}{3}...(i)$

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