If the area enclosed by the parabolas P₁:2y=5x² and P₂:x²-y+6=0 is equal to the area enclosed by P₁ and y=αx,α>0, then α³ is equal to ____________.

Solution

Maths sir · 4 steps

Find the first area as a plain number, then write the second area in terms of α and set them equal.

1. First area. P₁ is y = 5/2x² and P₂ is y = x² + 6. They meet where 5/2x² = x² + 6, so x² = 4 and x = ± 2. Between these, P₂ lies above P₁, so the area is ∫₋₂²(6 - 3/2x²)dx = 2(12 - 4) = 16.
2. Second area. The line y = α x meets P₁ where 5/2x² = α x, so x = 0 and x = 2α/5. Between these points the line lies above the parabola.

   ∫₀^(2α/5)(α x - 5/2x²)dx = 2α³/25 - 4α³/75 = 2α³/75

3. Equate. Set the two areas equal: 2α³/75 = 16, so α³ = 8 × 75.
4. So. α³ = **600**.

The common slip: 300. Integrating only from 0 to 2 for the first area gives 8 instead of 16, which leads to 300. The region is symmetric, so double it.

The idea to keep. Area between y = kx² and the line y = mx is always m³/6k². Remember it and questions like this take under a minute.

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