Let e₁ and e₂ be the eccentricities of the ellipse x²/b² + y²/25 = 1 and the hyperbola x²/16 - y²/b² = 1, respectively. If b < 5 and e₁e₂ = 1 , then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

Solution

Maths sir · 5 steps

Use e₁e₂ = 1 to find b first, and then the four foci sit right on the axes for you.

1. Both eccentricities. Since b < 5, the ellipse has its major axis along the y-axis, so e₁² = 1 - b²/25. For the hyperbola, e₂² = 1 + b²/16.

   (1 - b²/25)(1 + b²/16) = 1 ⇒ (25 - b²)(16 + b²) = 400

2. Solve for b. Expanding gives 400 + 9b² - b⁴ = 400, so b²(9 - b²) = 0. Since b ≠ 0, b² = 9. Then e₁ = 4/5 and e₂ = 5/4.
3. Locate the foci. The ellipse foci are (0, ± 5e₁) = (0, ± 4). The hyperbola foci are (± 4e₂, 0) = (± 5, 0).
4. New ellipse. An ellipse with axes along the coordinate axes through (± 5, 0) and (0, ± 4) is x²/25 + y²/16 = 1. These four foci are its vertices.
5. So. Its eccentricity is √(1 - 16/25) = 3/5, which is option (b).

The idea to keep. When points lie on the axes, they are the vertices of the new conic: the semi-axes come straight from them. Always check which axis is major before writing e.

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