Let the hyperbola H:x²/a² - y² = 1 and the ellipse E:3x² + 4y² = 12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If e_H and e_E are the eccentricities of H and E respectively, then the value of 12( e_H² + e_E² ) is equal to ___________.

Solution

Maths sir · 3 steps

Get the ellipse's latus rectum first, then use it to find a for the hyperbola.

1. Ellipse in standard form. Divide by 12: x²/4 + y²/3 = 1, so a² = 4 and b² = 3. Its latus rectum is 2b²/a = (2 × 3)/2 = 3, and e_E² = 1 - 3/4 = 1/4.
2. Match the latus recta. For H, b² = 1, so its latus rectum is 2b²/a = 2/a. Setting 2/a = 3 gives a = 2/3, so a² = 4/9.

   e_H² = 1 + b²/a² = 1 + 1/(4/9) = 1 + 9/4 = 13/4

3. So. 12(e_H² + e_E²) = 12(13/4 + 1/4) = 12 × 7/2 = 42.

The idea to keep. Latus rectum is 2b²/a for both ellipse and hyperbola. Only the eccentricity changes sign: 1 - b²/a² for the ellipse, 1 + b²/a² for the hyperbola.

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