Let PQ be a diameter of the circle x² + y² = 9. If α and β are the lengths of the perpendiculars from P and Q on the straight line,
x + y = 2 respectively, then the maximum value of αβ is _____.

Solution

Maths sir · 4 steps

P and Q are opposite ends of a diameter, so write both with one angle and multiply the two distances.

1. Parametrise the diameter. Take P = (3 cos θ, 3 sin θ). Since PQ is a diameter through the centre (0, 0), Q = (-3 cos θ, -3 sin θ).
2. Two distances. Let t = cos θ + sin θ. The distance from a point to x + y - 2 = 0 gives α = (|3t - 2|)/√2 and β = (|-3t - 2|)/√2 = (|3t + 2|)/√2.

   αβ = (|9t² - 4|)/2 = (|9(1 + sin 2θ) - 4|)/2 = (|5 + 9 sin 2θ|)/2

3. Find the maximum. Here t² = 1 + sin 2θ. As sin 2θ runs from -1 to 1, the quantity 5 + 9 sin 2θ runs from -4 to 14. Its largest absolute value is 14, at sin 2θ = 1.
4. So. The maximum value of αβ is 14/2 = 7.

The common slip: 14. Forgetting that each distance has √2 in the denominator gives |5 + 9 sin 2θ| with no division by 2, so the maximum comes out as 14 instead of 7.

The idea to keep. For the two ends of a diameter, use one angle θ and the opposite point. The product of the two distances becomes a difference of squares, which reduces to a single sin 2θ.

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