JEE-Mains 2014 (20/30)

Geometry Level 4

The locus of the foot of perpendicular drawn from the center of the ellipse x 2 + 3 y 2 = 6 x^2+3y^2=6 on any tangent to it is :

( x 2 − y 2 ) 2 = 6 x 2 − 2 y 2 (x^2-y^2)^2=6x^2-2y^2 ( x 2 − y 2 ) 2 = 6 x 2 + 2 y 2 (x^2-y^2)^2=6x^2+2y^2 ( x 2 + y 2 ) 2 = 6 x 2 + 2 y 2 (x^2+y^2)^2=6x^2+2y^2 ( x 2 + y 2 ) 2 = 6 x 2 − 2 y 2 (x^2+y^2)^2=6x^2-2y^2

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Rajen Kapur
Jun 15, 2015

Typo: In the answer, 's' is to be made 'x'.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Updated ! Thank you very much. :)

Sandeep Bhardwaj - 5 years, 12 months ago

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