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  • A point P moves so that the distance between the feet of perpendiculars drawn from it to the lines <img alt="\mathrm{a x^2+ 2 h x y+b y^2=0}" src="https://entrancecorner.oncodecogs.com/gif.lat

A point P 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 2k. Then the locus of P is \mathrm{\left(x^2+y^2\right)\left(h^2-a b\right)=k^2\left\{(a-b)^2+h^2\right\}}, where c= ?

Option: 1

1


Option: 2

3


Option: 3

4


Option: 4

2


Answers (1)

O, M, P, N are concyclic with centre \mathrm{O}^{\prime}\left(\mathrm{x}_1 / 2\right., \mathrm{y}_1 / 2 )and diameter OP,
Hence equation of circle will be (\mathrm{x}-0)\left(\mathrm{x}-\mathrm{x}_1\right)+(\mathrm{y}- 0) \left(\mathrm{y}-\mathrm{y}_1\right)=0
If \theta be the angle between the lines

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

\mathrm{\frac{2 \sqrt{\left(h^2-a b\right)}}{(a-b)^2+4 h^2}}  \mathrm{=\frac{\mathrm{k}}{\sqrt{\left(\frac{\mathrm{x}_1^2}{4}+\frac{\mathrm{y}_1^2}{4}\right)}}}

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

Hence locus of \mathrm{\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right) is \left(\mathrm{x}^2+\mathrm{y}^2\right)\left(\mathrm{h}^2-\mathrm{ab}\right)=\mathrm{k}^2\{(\mathrm{a}- b \left.)^2+4 h^2\right\}}

Posted by

Sumit Saini

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