Let an ellipse E:x²/a² + y²/b² = 1, a² > b², passes through ( √(3/2) ,1 ) and has eccentricity 1/√3. If a circle, centered at focus F(α, 0), α > 0, of E and radius 2/√3, intersects E at two points P and Q, then PQ² is equal to :

Solution

Maths sir · 5 steps

Fix the ellipse first; the radius is the semi-latus rectum, so P and Q are the latus rectum ends.

1. Find a and b. With e² = 1/3, b² = a²(1 - e²) = 2a²/3. Putting (√(3/2), 1) into the ellipse gives 3/2a² + 3/2a² = 1, so a² = 3 and b² = 2.
2. Locate the focus. ae = √3 × 1/√3 = 1, so the focus is F(1, 0) and α = 1.
3. Use focal distance. For any point (x, y) on the ellipse, its distance from F is a - ex. Set this equal to the radius 2/√3.

   √3 - x/√3 = 2/√3 ⇒ x = 1

4. Find P and Q. At x = 1, y² = 2(1 - 1/3) = 4/3, so P, Q = (1, ±2/√3). These are the ends of the latus rectum through F.
5. So. PQ = 4/√3, so PQ² = 16/3, which is option (c).

The idea to keep. When a circle is centred at a focus, use the focal distance a - ex instead of solving two quadratics. If the radius equals b²/a, the points are the latus rectum ends.

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