Question
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.
$${\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$
B.
$${\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} - 2{y^2}$$
C.
$${\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$
D.
$${\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} - 2{y^2}$$
Answer :
$${\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$
Solution :
Given equation of ellipse can be written as
$$\frac{{{x^2}}}{6} + \frac{{{y^2}}}{2} = 1\,\,\,\,\,\,\, \Rightarrow {a^2} = 6,\,\,\,{b^2} = 2$$
Now, equation of any variable tangent is
$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} .....({\text{i}})$$
where $$m$$ is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
$$y = \frac{{ - x}}{m}.....({\text{ii}})$$
Eliminating $$m,$$ we get
$$\eqalign{
& \left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2} \cr
& \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2} \cr
& \Rightarrow \boxed{{{\left( {{x^2} + {y^2}} \right)}^2} = 6{x^2} + 2{y^2}} \cr} $$