Let one focus of the hyperbola H: x²/a²-y²/b²=1 be at (√10, 0) and the corresponding directrix be x=9/√10. If e and l respectively are the eccentricity and the length of the latus rectum of H , then 9(e²+l) is equal to :

Solution

Maths sir · 4 steps

Focus gives ae, directrix gives a/e: multiply them and a² falls out at once.

1. Read off two facts. For x²/a²-y²/b²=1, the focus is at (ae, 0) and the corresponding directrix is x = a/e. So ae = √10 and a/e = 9/√10.
2. Find a and e. Multiplying the two gives a² = √10 × 9/√10 = 9, so a = 3. Then e = √10/3 and e² = 10/9.
3. Find b and l. For a hyperbola b² = a²(e² - 1) = 9 × 1/9 = 1. The latus rectum is l = 2b²/a = 2/3.

   9(e² + l) = 9(10/9 + 2/3) = 10 + 6 = 16

4. So. 9(e² + l) = 16, which is option (d).

The idea to keep. Whenever a focus and its directrix are both given, multiply ae by a/e to get a² and divide them to get e². For a hyperbola, use b² = a²(e² - 1).

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