If the foci of a hyperbola are same as that of the ellipse x²/9+y²/25=1 and the eccentricity of the hyperbola is 15/8 times the eccentricity of the ellipse, then the smaller focal distance of the point (√2, 14/3 √(2/5)) on the hyperbola, is equal to

Solution

Maths sir · 4 steps

The foci lie on the y-axis, so build a vertical hyperbola and use focal distance = |ey ± A|.

1. Read the ellipse. Here 25 > 9, so the major axis is along the y-axis with a = 5, b = 3. Then e = √(1 - 9/25) = 4/5 and the foci are (0, ± 4).
2. Build the hyperbola. Its eccentricity is 15/8 × 4/5 = 3/2. The foci (0, ± 4) are on the y-axis, so take y²/A² - x²/B² = 1 with Ae = 4. This gives A = 8/3 and B² = A²(e² - 1) = 64/9 × 5/4 = 80/9.
3. Check the point. With y² = 392/45 and x² = 2: y²/A² - x²/B² = 49/40 - 9/40 = 1. The point is on the hyperbola, and it lies on the upper branch.

   focal distances = |ey ± A|, ey = 3/2 × 14/3√(2/5) = 7√(2/5)

4. So. The two focal distances are 7√(2/5) ± 8/3. The smaller one is 7√(2/5) - 8/3 ≈ 1.76, which is option (c).

The idea to keep. First check which axis the foci are on. For a hyperbola with foci on the y-axis, the focal distances of a point are |ey ± A|. They differ by 2A, so the smaller one uses the minus sign.

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