Let y²=12 x be the parabola and S be its focus. Let P Q be a focal chord of the parabola such that (S P)(S Q)=147/4. Let C be the circle described taking P Q as a diameter. If the equation of a circle C is 64 x²+64 y²-α x-64 √3 y=β, then β-α is equal to .

Solution

Maths sir · 4 steps

Take the ends at t and -1/t, and let SP · SQ fix t²+1/t² for you.

1. Set up the chord. Here 4a = 12, so a = 3 and S = (3, 0). Take P = (3t², 6t). For a focal chord the other end is Q = (3/t², -6/t), with SP = 3(1+t²) and SQ = 3(1+1/t²).
2. Use the product. SP · SQ = 9(2 + t² + 1/t²) = 147/4. This gives t² + 1/t² = 49/12 - 2 = 25/12, so (t - 1/t)² = 1/12 and t - 1/t = ±1/2√3.
3. Diameter form. With ends (x₁, y₁) and (x₂, y₂), the circle is (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. Here x₁ x₂ = 9, y₁ y₂ = -36, x₁ + x₂ = 3 × 25/12 = 25/4 and y₁ + y₂ = 6(t - 1/t) = ±√3. The given -64√3 y term fixes the sign: y₁ + y₂ = √3.

   x² + y² - 25/4x - √3 y + (9 - 36) = 0 ⇒ 64x² + 64y² - 400x - 64√3 y = 1728

4. So. Matching with the given equation, α = 400 and β = 1728, so β - α = 1328.

The common slip: -2128. Leaving the constant on the left as -1728 and reading β = -1728 gives -2128. Move it to the right side first, so β = +1728.

The idea to keep. For a focal chord of y² = 4ax, the ends are t and -1/t. Then x₁x₂ = a² and y₁y₂ = -4a², so the circle on the chord as diameter comes out almost for free.

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