JEE Main 2021 · 27 August · evening · Single correct · +4 / −1
If two tangents drawn from a point P to the
parabola y² = 16(x - 3) are at right angles, then the locus of point P is :
- (a) x + 3 = 0
- (b) x + 1 = 0
- (c) x + 2 = 0
- (d) x + 4 = 0
Answer and solution
Answer (b) x + 1 = 0
Solution
Maths sir · 4 steps
Perpendicular tangents always meet on the directrix, so find the directrix of this shifted parabola.
- Shift the parabola. Put X = x - 3 and Y = y. The curve becomes Y² = 16X, which is Y² = 4aX with a = 4.
- Tangent through P. A tangent of slope m to Y² = 16X is Y = mX + 4/m. Let P = (h, k), so in shifted form P = (h - 3, k). Substituting and multiplying by m gives a quadratic in m.
(h-3)m² - km + 4 = 0
- Perpendicular condition. The two roots m₁, m₂ are the slopes of the two tangents from P. Perpendicular means m₁ m₂ = -1, so 4/(h-3) = -1, which gives h - 3 = -4 and h = -1.
- So. The locus of P is x = -1, that is, x + 1 = 0. This is the directrix X = -4 of the shifted parabola, written back in terms of x. The answer is (b).
The idea to keep. For any parabola, the locus of the point where perpendicular tangents meet is its directrix. For a shifted parabola, find the directrix in shifted coordinates first, then shift back.
Answer from the NTA final answer key. Solution by PrepWiser.