QJEEM14Maba8d: JEE Main 2014 Mathematics - Class 11 JEEM14Maba8d Solution: The locus of the foot of perpendicular drawn from the centre of …

Question

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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 :

Accepted Answer

Given

e{q^n}

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}

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