Question

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

• A. $(x^2+y^2{)}^2=5x^2+7y^2$
• B. $(x^2+y^2{)}^2=7x^2+5y^2$
• C. $(x^2+y^2{)}^2=x^2+3y^2$
• D. $(x^2+y^2{)}^2=3x^2+y^2$

Answer 1

The correct option is D $(x^2+y^2{)}^2=3x^2+y^2$

For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,

Centre is $(0,0)$ and any tangent is written in form $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$, then foot of perpendicular from center to this line is given by $(\frac{a{b}^2\cos \theta }{b^2{(\cos \theta )}^2+a^2{(\sin \theta )}^2},\frac{b{a}^2\sin \theta }{b^2{(\cos \theta )}^2+a^2{(\sin \theta )}^2})$
By eliminating $\theta$ from $x$ and $y$, we get locus of perpendicular as $a^2{x}^2+b^2{y}^2={(x^2+y^2)}^2$,
Now we get locus of perpendicular for above ellipse as $3x^2+y^2={(x^2+y^2)}^2$