If e₁ and e₂ are the eccentricities of the ellipse, x²/18 + y²/4 = 1 and the hyperbola, x²/9 - y²/4 = 1 respectively and (e₁, e₂) is a point on the ellipse, 15x² + 3y² = k, then k is equal to :

Solution

Maths sir · 3 steps

The equation only has x² and y², so find e₁² and e₂² and never take a square root.

1. Ellipse eccentricity. For x²/18 + y²/4 = 1, a² = 18 and b² = 4. So e₁² = 1 - b²/a² = 1 - 4/18 = 7/9.
2. Hyperbola eccentricity. For x²/9 - y²/4 = 1, a² = 9 and b² = 4. For a hyperbola the sign flips, so e₂² = 1 + b²/a² = 1 + 4/9 = 13/9.

   k = 15e₁² + 3e₂² = 15 × 7/9 + 3 × 13/9 = (105 + 39)/9 = 144/9

3. So. The point (e₁, e₂) lies on 15x² + 3y² = k, so k = 144/9 = 16, which is option (b).

The idea to keep. Ellipse: e² = 1 - b²/a². Hyperbola: e² = 1 + b²/a². When a point (e₁, e₂) goes into a conic in x² and y², work with e² directly.

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