Question

The locus of the foot of perpendicular drawn from the centre of the ellipse $ x^2 + 3y^2 = 6$ on any tangent to it is

• A. $ (x^2 + y^2)^2 = 6x^2 + 2y^2$
• B. $ (x^2 + y^2)^2 = 6x^2 - 2y^2$
• C. $(x^2 - y^2)^2 = 6x^2 + 2y^2$
• D. $(x^2 - y^2)^2 = 6x^2 - 2y^2$

Answer 1

The Correct Option is A

$ Equation \, \, of \, ellipse \, is x^2+ 3 y^2 = 6 or \frac{x^2}{6} + \frac{y^2}{2} = 1 $
Equation of the t. angent. is $ \frac{xcos \theta}{a } + \frac{y sin \theta}{b } = 1 $
Let (6, k) be any point on the locus.
$ \therefore $ $\hspace20mm$ $\frac{ h}{a} cos \theta +\frac{ k}{b} sin \theta =1$ ......(i)
Slope of the tangent line is $\frac{-b}{a} cot \theta$
Slope of perpendicular drawn from centre (0,0) to (6, k) is k/h
Since, both the lines are perpendicular.
$\therefore \, \, \, \, \, \, \, \, \, \bigg(\frac{k}{h} \bigg) \times \bigg( -\frac{b}{a}cot \theta \bigg) = -1 $
$\Rightarrow \, \, \, \, \frac{ cos \theta}{ha} = \frac{ sin \theta}{ kb} = \alpha \, \, \, \, \, [say] $
$\Rightarrow cos \theta = \alpha h a $
$\hspace15mm$ $ sin \theta = \alpha k b $
From E1 $ \frac{h}{a}(\alpha h a) + \frac{k}{b}( \alpha k b ) = 1$
$\Rightarrow \, \, \, \, \, \, \, \, \, h^2 \alpha + k^2 \alpha = 1 $
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \alpha= \frac{1}{h^2 +k^2} $
Also , $ sin ^2 \alpha +cos ^2 \alpha =1 $
$\Rightarrow \, \, \, \, \, \, ( \alpha k b)^2+( \alpha h a)^2 =1 \Rightarrow \alpha^2 k^2 b^2 + \alpha^2 h^2 a^2 =1$
$\Rightarrow \frac{k^2 b^2 }{(h^2 + k^2)^2} + \frac{h^2 a^2} {(h^2+ k^2)^2} =1$
$\Rightarrow \frac{2 k^2 }{(h^2 + k^2)^2} + \frac{ 6 h^2 } {(h^2+ k^2)^2} =1 \, \, \, \, \, \, [ \therefore a^2= 6 , b^2 =2 ] $
$\Rightarrow 6x^2 +2 y^2 = (x^2 +y^2)^2 $
$\hspace25mm$ [replacing k by y and h by x]