JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

If polar of

y^2=4 a x

is always touching the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

, then locus of the pole is

View answer and explanation

Correct option: A

4 a^2 x^2-b^2 y^2=4 a^4

Option A (

4 a^2 x^2-b^2 y^2=4 a^4

) is correct because the polar of the parabola touches the ellipse, so the condition for tangency can be used to find the locus of the pole.

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