question_answer

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

A) \[{{({{x}^{2}}-{{y}^{2}})}^{2}}=6{{x}^{2}}+2{{y}^{2}}\]

B) \[{{({{x}^{2}}-{{y}^{2}})}^{2}}=6{{x}^{2}}-2{{y}^{2}}\]

C) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=6{{x}^{2}}+2{{y}^{2}}\]

D) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=6{{x}^{2}}-2{{y}^{2}}\]

Correct Answer: C

Solution :

Equation of ellipse, \[{{x}^{2}}+3{{y}^{2}}=6\] Equation of tangent of ellipse with slope m is

\[y=mx\pm \sqrt{6{{m}^{2}}+2}\]

\[m=-\frac{h}{k}\]

and tangent passes through \[(h,k)\]

\[\therefore \]\[k=h\left( -\frac{h}{k} \right)\pm \sqrt{\frac{6{{h}^{2}}}{{{k}^{2}}}+2}\]\[\Rightarrow \]\[{{k}^{2}}+{{h}^{2}}=\pm \sqrt{6{{h}^{2}}+2{{k}^{2}}}\]\[\Rightarrow \]\[{{({{h}^{2}}+{{k}^{2}})}^{2}}=6{{h}^{2}}+2{{k}^{2}}\]So, locus is \[({{x}^{2}}+{{y}^{2}})=6{{x}^{2}}+2{{y}^{2}}.\]