JEE Main | 2014EllipseHard
Question
The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it is
Options
A.(x2 + y2)2 = 6x2 + 2y2
B.(x2 + y2)2 = 6x2 - 2y2
C.(x2 - y2)2 = 6x2 + 2y2
D.(x2 - y2)2 = 6x2 - 2y2
Solution
Here ellipse is 
= 1, where a2 = 6, b2 = 2
Now, equation of any variable tangent is
y = mx±
....(i)
where m is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
y =
....(ii)
Eliminating m, we get
(x2 + y2)2 = a2x2 + b2y2
⇒
= 1, where a2 = 6, b2 = 2Now, equation of any variable tangent is
y = mx±
....(i)where m is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
y =
....(ii)Eliminating m, we get
(x2 + y2)2 = a2x2 + b2y2
⇒
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