Let the foci of a hyperbola H coincide with the foci of the ellipse E: (x-1)²/100+(y-1)²/75=1 and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is α and the length of its conjugate axis is β, then 3 α²+2 β² is equal to

Solution

Maths sir · 4 steps

The ellipse gives you two things for free: the focal distance 5 and the eccentricity to flip.

1. Read the ellipse. The centre is (1, 1), with a² = 100 and b² = 75. So e² = 1 - 75/100 = 1/4, which gives e = 1/2. The foci lie at distance ae = 10 × 1/2 = 5 from the centre, along the line y = 1.
2. Hyperbola's a. H has the same foci, so its centre is also (1, 1) and its focal distance is also 5. Its eccentricity is the reciprocal, e_H = 2. Then a_H × 2 = 5, so a_H = 5/2.

   b_H² = a_H²(e_H² - 1) = 25/4 × 3 = 75/4

3. Axis lengths. The transverse axis is α = 2a_H = 5, so α² = 25. The conjugate axis is β = 2b_H, so β² = 4b_H² = 75.
4. So. 3α² + 2β² = 3 × 25 + 2 × 75 = 75 + 150 = 225, which is option (a).

The idea to keep. Shared foci means the same centre and the same focal distance ae. Fix ae first, then use e to get a. For a hyperbola, b² = a²(e² - 1), and axis lengths are 2a and 2b.

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