Consider the circle C: x²+y²=4 and the parabola P: y²=8 x. If the set of all values of α, for which three chords of the circle C on three distinct lines passing through the point (α, 0) are bisected by the parabola P is the interval (p, q), then (2 q-p)² is equal to __________.

Solution

Maths sir · 4 steps

Take the midpoint as (2t², 4t), write its chord as T = S₁, and make it pass through (α, 0).

1. Chord from midpoint. Any point of P is (2t², 4t). The chord of x²+y²=4 with midpoint (h, k) is T = S₁, that is hx + ky = h² + k². So the chord bisected at (2t², 4t) is 2t²x + 4ty = 4t⁴ + 16t².

   Through (α, 0): 2t²α = 4t⁴ + 16t² ⇒ t = 0 or α = 2t² + 8

2. Count the lines. t = 0 gives the midpoint (0, 0). Every diameter is bisected there, and the one through (α, 0) is the x-axis, which is one line. The other two lines need t² = (α - 8)/2 > 0, giving t = ±√((α-8)/2), so α > 8.
3. Midpoint inside circle. A real chord needs its midpoint strictly inside the circle: 4t⁴ + 16t² < 4, so t⁴ + 4t² - 1 < 0. This gives t² < √5 - 2, and then α = 2t² + 8 < 4 + 2√5.
4. So. α ∈ (8, 4 + 2√5), so p = 8 and q = 4 + 2√5. Then 2q - p = 4√5, and (2q - p)² = 80.

The idea to keep. For chords bisected at a point, always use T = S₁. Then add the hidden condition: the midpoint must lie strictly inside the circle, or the chord does not exist.

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