The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6 on any tangent to it is :
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Official solution
Given eqn of ellipse can be written as
\[{{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2\]
Now, equation of any variable tangent is
\[y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)\]
where m is slope of the tangent
So, equation of perpendicular line drawn-
from center to tangent is
\[y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\]
Eliminating m, we get
\[\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}\]
\[ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}\]
\[ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}\]