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JEE Main 2014

JEE Main 2014 Mathematics question on Conic Sections

The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6x^2 + 3y^2 = 6 on any tangent to it is
  1. A.
    (x2−y2)2=6x2+2y2(x^2 - y^2)^2 = 6x^2 + 2y^2
  2. B.
    (x2−y2)2=6x2−2y2(x^2 - y^2)^2 = 6x^2 - 2y^2
  3. C.
    (x2+y2)2=6x2+2y2(x^2 + y^2)^2 = 6x^2 + 2y^2
  4. D.
    (x2+y2)2=6x2−2y2(x^2 + y^2)^2 = 6x^2 - 2y^2
Show answer and solution

Correct option: C

Solution

The ellipse is x²/6 + y²/2 = 1, so a² = 6, b² = 2. Any tangent to this ellipse can be written as y = mx + √(6m² + 2).

Let the foot of perpendicular from centre (0,0) to this tangent be (h, k). The tangent has slope m, and since the perpendicular from origin to the tangent has slope -1/m, we get k/h = -1/m, so m = -h/k.

Also, the foot (h, k) lies on the tangent: k = mh + √(6m² + 2), which gives (k - mh)² = 6m² + 2.

Since k - mh = k - (-h/k)h = (k² + h²)/k, we have:
(h² + k²)²/k² = 6(h²/k²) + 2

Multiply both sides by k²:
(h² + k²)² = 6h² + 2k²

Replacing (h, k) with (x, y): (x² + y²)² = 6x² + 2y², which is option 3.

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