Let the tangents at the points P and Q on the ellipse x²/2+y²/4=1 meet at the point R(√2, 2 √2-2). If S is the focus of the ellipse on its negative major axis, then SP²+SQ² is equal to ___________.

Solution

Maths sir · 5 steps

Find P and Q from the chord of contact of R, then measure both distances from the lower focus.

1. Locate the focus. Here 4 > 2, so the major axis is the y-axis, with b² = 4 and a² = 2. Then e² = 1 - 2/4 = 1/2, so the foci are (0, ± be) = (0, ±√2). The focus on the negative major axis is S(0, -√2).
2. Chord of contact. P and Q lie on the chord of contact of R(√2, 2√2-2), which is (√2 x)/2 + ((2√2-2) y)/4 = 1. Put x = √2 cos θ, y = 2 sin θ and simplify.

   cos θ + (√2-1)sin θ = 1 ⇒ cos(θ - π/8) = cos π/8

3. Find P and Q. This uses tan π/8 = √2-1, and it gives θ = 0 or θ = π/4. So P = (√2, 0) and Q = (1, √2). Check: the tangent at P is x = √2, which does pass through R.
4. Measure from S. SP² = (√2)² + (√2)² = 4. SQ² = 1² + (2√2)² = 1 + 8 = 9.
5. So. SP² + SQ² = 4 + 9 = 13.

The common slip: 5. Using the upper focus (0, √2) instead of the lower one gives 4 + 1 = 5, not 13.

The idea to keep. Tangents from an outside point touch the ellipse where the chord of contact meets it. Before using any focus, check which denominator is larger: that tells you the major axis.

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