The line x=8 is the directrix of the ellipse E:x²/a²+y²/b²=1 with the corresponding focus (2,0). If the tangent to E at the point P in the first quadrant passes through the point (0,4√3) and intersects the x-axis at Q, then (3PQ)² is equal to ____________.

Solution

Maths sir · 4 steps

Focus and directrix fix a and e at once; then the outside point pins down P through the tangent equation.

1. Find the ellipse. The focus is (ae, 0) and the directrix is x = a/e, so ae = 2 and a/e = 8. Multiplying gives a² = 16, so a = 4 and e = 1/2. Then b² = a²(1 - e²) = 16 × 3/4 = 12.

   (x x₁)/16 + (y y₁)/12 = 1

2. Locate P. This is the tangent at P(x₁, y₁). It passes through (0, 4√3), so (4√3 y₁)/12 = 1, which gives y₁ = √3. On the ellipse, x₁²/16 = 1 - 3/12 = 3/4, so x₁ = 2√3 because P is in the first quadrant.
3. Find Q and PQ. Put y = 0 in the tangent: x = 16/x₁ = 8/√3, so Q = (8/√3, 0). Then PQ² = (8/√3 - 6/√3)² + (√3)² = 4/3 + 3 = 13/3.
4. So. (3PQ)² = 9 × 13/3 = 39.

The common slip: 13. Squaring only PQ and not the 3 gives 3 × 13/3 = 13. The square applies to 3PQ as a whole, so multiply PQ² by 9.

The idea to keep. For an ellipse, focus (ae, 0) times directrix a/e gives a² directly. When a tangent passes through a known point, put that point into xx₁/a² + yy₁/b² = 1 to find the point of contact.

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