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A point moves so that the distance between the feet of perpendiculars drawn from it to the lines $\mathrm{a x^2+2 h x y+b y^2=0}$ is a constant $\mathrm{ 2 k}$ . The equation of its locus is $\mathrm{ \left(x^2+y^2\right)\left(h^2-a b\right)=4 c^2\left\{(a-b)^2+4 h^2\right\}}$ where $\mathrm{ c=}$

Option: 1
$\mathrm{K}$

Option: 2
$\mathrm{\frac{k}{2}}$

Option: 3
$\mathrm{2K}$

Option: 4
$\mathrm{\frac{2k}{3}}$

Answers (1)

best_answer

$\mathrm{O}, \mathrm{M}, \mathrm{P}, \mathrm{N}$ are concyclic with center $\mathrm{O}^{\prime}\left(\mathrm{x}_1 / 2\right., \mathrm{y}_1 / 2 )$ and diameter $\mathrm{OP}$ ,
Hence equation of circle will be

$\mathrm{(x-0)\left(x-x_1\right)+(y-0)\left(y-y_1\right)=0}$

If $\theta$ be the angle between the lines

$\mathrm{ \therefore \quad \tan \theta=\frac{2 \sqrt{\left(h^2-a b\right)}}{(a+b)} }$

$\mathrm{ \Rightarrow \quad \sin \theta=\frac{2 \sqrt{\left(h^2-a b\right)}}{\sqrt{(a-b)^2+4 h^2}}=\frac{k}{\sqrt{\left(\frac{x_1^2}{4}+\frac{y_1^2}{4}\right)}} }$
$\mathrm{ \Rightarrow \quad \sqrt{\left(x_1^2+y_1^2\right.} \sqrt{\left(h^2-a b\right)}=k \sqrt{(a-b)^2+4 h^2} }$

$\mathrm{ \Rightarrow \quad\left(x_1^2+y_1^2\right)\left(h^2-a b\right)=k^2\left\{(a-b)^2+4 h^2\right\}}$

Hence locus of $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is

$\mathrm{\left(x^2+y^2\right)\left(h^2-a b\right)=k^2\left\{(a-b)^2+4 h^2\right\}}$

Hence option 2 is correct.

Posted by

avinash.dongre

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