Let the ellipse 3x² + py² = 4 pass through the centre C of the circle x² + y² - 2x - 4y - 11 = 0 of radius r. Let f₁, f₂ be the focal distances of the point C on the ellipse. Then 6f₁f₂ - r is equal to

Solution

Maths sir · 4 steps

Use the point C to find p, then notice the major axis is vertical, so focal distances are 4 ± ey.

1. Centre and radius. Compare x² + y² - 2x - 4y - 11 = 0 with x² + y² + 2gx + 2fy + c = 0. The centre is C(1, 2) and r = √(1 + 4 + 11) = 4.
2. Find p. C lies on the ellipse, so 3(1)² + p(2)² = 4, which gives p = 1/4. The ellipse is x²/(4/3) + y²/16 = 1.
3. Vertical major axis. Since 16 > 4/3, the semi-major axis is 4, along the y-axis. Then e² = 1 - (4/3)/16 = 11/12. The foci are (0, ± 4e), and for a point with ordinate y the focal distances are 4 + ey and 4 - ey.

   f₁ f₂ = (4 + 2e)(4 - 2e) = 16 - 4e² = 16 - 11/3 = 37/3

4. So. 6f₁f₂ - r = 6 × 37/3 - 4 = 74 - 4 = 70, which is option (c).

The idea to keep. Focal distances are (semi-major axis) ± e × (coordinate along the major axis). Always check which denominator is bigger before you pick x or y.

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