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 C $(x^2+y^2{)}^2=6x^2+2y^2$

Equation of ellipse is $x^2+3y^2=6or\frac{x^2}{6}+\frac{y^2}{2}=1$
Equation of the tangent is $\frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$
Let (h,k) be any point on the locus.
$\frac{h}{a}cos\theta +\frac{k}{b}sin\theta =1-------\left(i\right)$
Slope of the tangent line is $-\frac{b}{a}cot\theta$
slope of perpendicular drawn form centre (0,0) to (h,k) is $\frac{k}{h}$
Since both the line are perpendicular.

$\therefore \left(\frac{k}{h}\right)\times (-\frac{b}{a}cot\theta )=-1\Rightarrow \frac{cos\theta }{ha}=\frac{sin\theta }{kb}=\alpha \Rightarrow cos\theta =\alpha ha,andsin\theta =\alpha kbFromeq\left(i\right)\frac{h}{a}\left(\alpha ha\right)+\frac{k}{b}\left(\alpha kb\right)=1\Rightarrow \alpha =\frac{1}{h^2+k^2}Alsosi{n}^2\theta +co{s}^2\theta =1\Rightarrow (\alpha kb{)}^2+(\alpha ha{)}^2=1\Rightarrow a^2{k}^2{b}^2+a^2{h}^2{a}^2=1\Rightarrow \frac{k^2{b}^2}{(h^2+k^2{)}^2}+\frac{h^2{a}^2}{(h^2+k^2{)}^2}=1\Rightarrow \frac{2k^2}{(h^2+k^2{)}^2}+\frac{6h^2}{(h^2+k^2{)}^2}=1\Rightarrow 6x^2+2y^2=(x^2+y^2{)}^2$