The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2y²=kx and ky²=2(y-x) is maximum, is equal to :

Solution

Maths sir · 4 steps

Both curves are easy as x in terms of y, so integrate along y and the area becomes a function of k.

1. Write x in y. The first curve gives x = 2y²/k and the second gives x = y - k/2y². Their difference is y - a y², where a = k/2 + 2/k.
2. Find the meeting points. Setting y - a y² = 0 gives y = 0 or y = 1/a. These are the limits of integration along y.

   Area = | ∫₀^(1/a) (y - a y²) dy | = 1/2a² - 1/3a² = 1/6a²

3. Make the area largest. The area is largest when |a| is smallest. By AM-GM, |k/2 + 2/k| = |k|/2 + 2/|k| ≥ 2, with equality only when |k| = 2. That gives k = 2 or k = -2, and the maximum area is 1/24.
4. So. The sum of squares is 2² + (-2)² = 4 + 4 = 8.

The common slip: 4. Taking only k = 2 and forgetting k = -2 gives 4, not 8. The area depends on a², so both signs give the same maximum.

The idea to keep. When the area comes out as a function of a parameter, look for a form like t + c/t and apply AM-GM to |t|. Always check both signs of the parameter.

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