Let the foci of the ellipse x²/16+y²/7=1 and the hyperbola x²/144-y²/α=1/25 coincide. Then the length of the latus rectum of the hyperbola is :

Solution

Maths sir · 4 steps

First make the hyperbola's right side equal to 1, then borrow c = 3 from the ellipse to find b².

1. Foci of ellipse. For the ellipse, a² = 16 and b² = 7, so c² = a² - b² = 9. The foci are (± 3, 0).
2. Standard form. Multiply the hyperbola by 25 to get x²/(144/25) - y²/(α/25) = 1. So a² = 144/25, which gives a = 12/5, and b² = α/25.
3. Match the foci. For a hyperbola c² = a² + b², and c = 3 here. So b² = 9 - 144/25 = 81/25.

   Latus rectum = 2b²/a = (2 × 81/25)/(12/5) = 162/25 × 5/12 = 27/10

4. So. The latus rectum of the hyperbola is 27/10, which is option (d).

The idea to keep. Always bring a conic to the form with 1 on the right before reading off a² and b². Ellipse: c² = a² - b². Hyperbola: c² = a² + b².

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