Let C be the largest circle centred at (2, 0) and inscribed in the ellipse x²/36 + y²/16 = 1. If (1, α) lies on C, then 10 α² is equal to ____________

Solution

Maths sir · 4 steps

The largest circle touches the ellipse at its nearest point, so find the minimum distance from (2, 0).

1. Radius equals distance. A circle centred at (2, 0) stays inside the ellipse only while its radius is no more than the distance to the nearest point of the ellipse. So the largest radius is that minimum distance.
2. Distance as function. Take a point (6 cos θ, 4 sin θ) on the ellipse and let c = cos θ. Its squared distance from (2, 0) is (6c - 2)² + 16(1 - c²).

   d² = 20c² - 24c + 20

3. Minimise. This quadratic is smallest at c = 24/40 = 0.6, which lies in [-1, 1]. Then d² = 20(0.36) - 24(0.6) + 20 = 12.8, so r² = 12.8 and the circle C is (x - 2)² + y² = 12.8.
4. So. Put (1, α) into C: 1 + α² = 12.8, so α² = 11.8 and 10α² = 118.

The common slip: 150. Taking the radius as the distance to the nearest vertex, 6 - 2 = 4, gives α² = 15 and 150. The circle would then cross the ellipse near (3.6, ± 3.2).

The idea to keep. For the biggest circle inside a curve with a fixed centre, the radius is the shortest distance from the centre to the curve. Parametrise the curve and minimise the squared distance.

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