Let a common tangent to the curves y² = 4x and (x - 4)² + y² = 16 touch the curves at the points P and Q. Then (PQ)² is equal to __________.

Solution

Maths sir · 5 steps

Get P from the parabola's slope form. Then PQ is just the tangent length from P to the circle.

1. Tangent to parabola. For y² = 4x we have a = 1, so every tangent can be written as y = mx + 1/m. It touches the parabola at (1/m², 2/m).
2. Touch the circle. The circle has centre C(4, 0) and radius 4. The line mx - y + 1/m = 0 is tangent to it when its distance from C equals 4.

   (|4m + 1/m|)/√(1+m²) = 4 ⇒ 16m² + 8 + 1/m² = 16 + 16m² ⇒ m² = 1/8

3. Find P. With m² = 1/8, the contact point is P = (8, ± 4√2). The y-axis is also a common tangent, but it touches both curves at the origin, so it gives no segment and is set aside.
4. Tangent length. The line PQ touches the circle at Q, so PQ is the tangent length from P. Hence PQ² = CP² - r² = (8-4)² + (4√2)² - 16 = 16 + 32 - 16.
5. So. PQ² = 32.

The common slip: 48. Forgetting to subtract r² = 16 gives CP² = 48, not 32. CP goes to the centre, while PQ stops at the circle.

The idea to keep. For a common tangent, use the parabola's slope form y = mx + a/m and apply the circle's distance condition. Once you have one contact point, the other piece is a tangent length: CP² - r².

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