If one end of a focal chord AB of the parabola y² = 8x is at A( 1/2, - 2 ), then the equation of the tangent to it at B is :

Solution

Maths sir · 4 steps

Write both ends in parameter form; on a focal chord t₁ t₂ = -1, so A gives you B at once.

1. Parameter form. Compare y² = 8x with y² = 4ax to get a = 2. Any point on it is (2t², 4t).
2. Find t for A. For A(1/2, -2), 4t₁ = -2 gives t₁ = -1/2. Check: 2t₁² = 1/2, which matches.
3. Locate B. For the ends of a focal chord, t₁ t₂ = -1, so t₂ = 2. Then B = (2 × 4, 4 × 2) = (8, 8).

   yy₁ = 2a(x + x₁) ⇒ 8y = 4(x + 8) ⇒ 2y = x + 8

4. So. The tangent at B is x - 2y + 8 = 0, which is option (b). As a check, it meets the tangent at A, 2x + y + 1 = 0, at (-2, 3), a point on the directrix x = -2.

The idea to keep. For a focal chord of y² = 4ax, the parameters of the two ends satisfy t₁ t₂ = -1. The tangents at the two ends are perpendicular and meet on the directrix.

Answer from the NTA final answer key. Solution by PrepWiser.