Question

Two lines are intersecting at origin and making equal angle $(\ne 45°)$ with $x-$axis. Locus of point whose sum of squares of its distances from these two lines is constant, will be

• A. $x^2+y^2=(ab{)}^2$, where $\frac{b}{a}=\tan \theta$,
  $\theta$ angle made by lines to $x-$axis
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $\frac{a}{b}=\tan \theta$,
  $\theta$ angle made by lines to $x-$axis
• C. $x^2+y^2=(ab{)}^2$, where $\frac{a}{b}=\tan \theta$,
  $\theta$ angle made by lines to $x-$axis
• D. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $\frac{b}{a}=\tan \theta$,
  $\theta$ angle made by lines to $x-$axis

Answer 1

The correct option is D $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $\frac{b}{a}=\tan \theta$,

$\theta$ angle made by lines to $x-$axis

Let two intersecting lines $OA$ and $OB$, intersect at origin $O$ and makes equal angle $\theta$ with $x-$axis
$\Rightarrow \angle XOA=\angle XOB=\theta$.

Equation of straight lines $OA$ and $OB$ are
$y=x\tan \theta$ and $y=-x\tan \theta$
or $x\sin \theta -y\cos \theta =0\dots \left(i\right)$
and $x\sin \theta +y\cos \theta =0\dots \left(ii\right)$

Let $P(\alpha ,\beta )$ is the point whose locus is to be determined
Given,
$(PM{)}^2+(PN{)}^2=2k^2$ (say)
$\Rightarrow \frac{(\alpha \sin \theta +\beta \cos \theta {)}^2}{\sin {\theta }^2+\cos {\theta }^2}+\frac{(\alpha \sin \theta -\beta \cos \theta {)}^2}{\sin {\theta }^2+\cos {\theta }^2}=2k^2$
$\Rightarrow 2{\alpha }^2{\sin }^2\theta +2{\beta }^2{\cos }^2\theta =2k^2$
$\Rightarrow {\alpha }^2{\sin }^2\theta +{\beta }^2{\cos }^2\theta =k^2$
$\Rightarrow \frac{{\alpha }^2}{(k\text{cosec}\theta {)}^2}+\frac{{\beta }^2}{(k\sec \theta {)}^2}=1$
Hence required locus is $\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
where $\frac{b}{a}=\tan \theta ,\theta$ is angle made by lines with $x-$axis