Let H be the hyperbola, whose foci are (1 ± √2, 0) and eccentricity is √2. Then the length of its latus rectum is :

Solution

Maths sir · 3 steps

The foci are not centred at the origin: find the centre first, then ae gives you a straight away.

1. Find the centre. The centre is the midpoint of the foci (1+√2, 0) and (1-√2, 0), which is (1, 0). Each focus is √2 away from the centre, so ae = √2.
2. Get a and b. With e = √2, ae = √2 gives a = 1. Then b² = a²(e² - 1) = 1 × (2 - 1) = 1.

   Latus rectum = 2b²/a = (2 × 1)/1 = 2

3. So. The hyperbola is (x-1)² - y² = 1, and its latus rectum has length 2, option (c).

The idea to keep. Shifting the centre does not change a, b or e. Take ae as half the distance between the foci, never as a focus coordinate, then use b² = a²(e²-1) and 2b²/a.

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