Let P((2 √3)/√7, 6/√7), Q, R and S be four points on the ellipse 9 x²+4 y²=36. Let PQ and RS be mutually perpendicular and pass through the origin. If 1/(P Q)²+1/(R S)²=p/q, where p and q are coprime, then p+q is equal to :

Solution

Maths sir · 4 steps

Both chords pass through the centre, so each is twice a radius; use polar form of the ellipse.

1. Chords are diameters. The ellipse is x²/4 + y²/9 = 1, centred at the origin. A chord through the centre is bisected there, so PQ = 2 OP and RS = 2 OR.
2. Polar form. Put a point at distance r from O in direction θ: (r cos θ, r sin θ). Substituting into the ellipse gives 1/r² = (cos² θ)/4 + (sin² θ)/9. For R, the direction is θ + 90°, so cos and sin swap places.

   1/OP² + 1/OR² = (cos² θ + sin² θ)/4 + (sin² θ + cos² θ)/9 = 1/4 + 1/9 = 13/36

3. Check with P. OP² = 12/7 + 36/7 = 48/7, so 1/OP² = 21/144. OP has slope √3, so θ = 60°, and at 150° we get 1/OR² = 3/16 + 1/36 = 31/144. The sum is 52/144 = 13/36, as expected.
4. So. 1/PQ² + 1/RS² = 1/4 × 13/36 = 13/144. Then p = 13, q = 144, and p + q = 157, option (d).

The idea to keep. For any two perpendicular semi-diameters of an ellipse, 1/r₁² + 1/r₂² = 1/a² + 1/b². The point P is given only to fix the picture; the sum does not depend on it.

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