JEE PYQ: Ellipse Question 11
Question 11 - 2020 (06 Sep Shift 1)
Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x^2}{4} + \frac{y^2}{2} = 1$ from any of its foci?
(a) $(-2, \sqrt{3})$ (b) $(-1, \sqrt{2})$ (c) $(-1, \sqrt{3})$ (d) $(1, 2)$
Show Answer
Answer: (c) $(-1, \sqrt{3})$
Solution
The locus of the feet of the perpendicular from foci to any tangent of the ellipse is the auxiliary circle $x^2 + y^2 = a^2 = 4$
$(-1, \sqrt{3})$: $1 + 3 = 4$ satisfies the equation.
BETA
