Let P(α, β) be a point on the parabola y²=4 x. If P also lies on the chord of the parabola x²=8 y whose mid point is (1, 5/4), then (α-28)(β-8) is equal to _________.

Solution

Maths sir · 4 steps

Find the chord from its midpoint first. Then P is where that line meets the other parabola.

1. Chord by midpoint. For x² = 8y, a chord whose midpoint is (x₁, y₁) is given by T = S₁, that is x x₁ - 4(y + y₁) = x₁² - 8y₁. Put in (x₁, y₁) = (1, 5/4).

   x - 4(y + 5/4) = 1 - 10 ⇒ x - 4y + 4 = 0

2. Meet the parabola. P(α, β) is on this line, so α = 4β - 4. It is also on y² = 4x, so β² = 4(4β - 4), which gives β² - 16β + 16 = 0.
3. Solve for both. β = 8 ± 4√3, so β - 8 = ± 4√3. Then α - 28 = 4β - 32 = 4(β - 8) = ± 16√3, with the same sign as β - 8.
4. So. (α - 28)(β - 8) = 4(β - 8)² = 4 × 48 = 192, for either point.

The common slip: -192. Pairing α = 28 + 16√3 with β = 8 - 4√3 mixes the two intersection points and gives -192. Both shifts must carry the same sign because α - 28 = 4(β - 8).

The idea to keep. Whenever a chord is fixed by its midpoint, write it as T = S₁. Then swap one variable using the line and you get a quadratic for the intersection points.

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