Let R be the focus of the parabola y²=20 x and the line y=m x+c intersect the parabola at two points P and Q.
Let the point G(10,10) be the centroid of the triangle P Q R. If c-m=6, then (P Q)² is :

Solution

Maths sir · 5 steps

Write P and Q in parametric form and let the centroid hand you t₁ + t₂ and t₁² + t₂².

1. Set up points. For y² = 20x, 4a = 20, so a = 5 and the focus is R(5, 0). Take P(5t₁², 10t₁) and Q(5t₂², 10t₂).
2. Use the centroid. The y-coordinate gives (10t₁ + 10t₂ + 0)/3 = 10, so t₁ + t₂ = 3. The x-coordinate gives (5t₁² + 5t₂² + 5)/3 = 10, so t₁² + t₂² = 5.

   t₁ t₂ = ((t₁ + t₂)² - (t₁² + t₂²))/2 = (9 - 5)/2 = 2

3. Find the points. With sum 3 and product 2, the parameters are 1 and 2. So P(5, 10) and Q(20, 20).
4. Check the line. The slope of PQ is m = 10/15 = 2/3. From 10 = 2/3 × 5 + c, c = 20/3, so c - m = 18/3 = 6, as given.
5. So. (PQ)² = (20 - 5)² + (20 - 10)² = 225 + 100 = 325, which is option (b).

The idea to keep. When points lie on a parabola, write them with parameters. A centroid or midpoint condition then becomes a sum and a sum of squares of the parameters, and the product follows from those two.

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