An ellipse E: x²/a²+y²/b²=1 passes through the vertices of the hyperbola H: x²/49-y²/64=-1. Let the major and minor axes of the ellipse E coincide with the transverse and conjugate axes of the hyperbola H, respectively. Let the product of the eccentricities of E and H be 1/2. If l is the length of the latus rectum of the ellipse E, then the value of 113 l is equal to _____________.

Solution

Maths sir · 4 steps

The minus sign flips this hyperbola: its vertices sit on the y-axis, so the ellipse is vertical too.

1. Read the hyperbola. Multiply by -1 to get y²/64 - x²/49 = 1. The transverse axis is the y-axis, so the vertices are (0, ± 8). Its eccentricity is e_H = √(1 + 49/64) = √113/8.
2. Fix the ellipse. The major axis of E lies along the y-axis and E passes through (0, ± 8). So b = 8 is the semi-major axis and a < b. The eccentricity of E is then e_E = √(1 - a²/b²).
3. Use the product. e_E × √113/8 = 1/2 gives e_E = 4/√113, so e_E² = 16/113. Then a²/64 = 1 - 16/113 = 97/113, which gives a² = (64 × 97)/113.

   l = 2a²/b = (2 × 64 × 97)/(113 × 8) = 1552/113

4. So. The major axis is along y, so the latus rectum uses the minor semi-axis squared over the major one: l = 1552/113. Hence 113 l = 1552.

The common slip: 1612. Writing e_H = √(1 + 64/49), as if the hyperbola were horizontal, gives e_H = √113/7 and leads to 1612, not 1552.

The idea to keep. Before using any formula, find which axis is the transverse or major one. For a vertical conic, swap the roles of a and b in the eccentricity and latus rectum formulas.

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