Let the foci and length of the latus rectum of an ellipse x²/a²+y²/b²=1, a>b b e( ± 5,0) and √50, respectively. Then, the square of the eccentricity of the hyperbola x²/b²-y²/(a² b²)=1 equals

Solution

Maths sir · 4 steps

For this hyperbola e² = 1 + a², so all you need from the ellipse is a².

1. Read the ellipse. The foci (± 5, 0) give a² - b² = 25. The latus rectum gives 2b²/a = √50 = 5√2, so b² = 5√2/2a.
2. Solve for a. Substituting gives a² - 5√2/2a - 25 = 0. The positive root is a = 5√2, so a² = 50 and b² = 50 - 25 = 25.

   e_H² = 1 + (a² b²)/b² = 1 + a²

3. Hyperbola eccentricity. In x²/b² - y²/a²b² = 1, the denominator under x² is b² = 25 and the denominator under y² is a²b² = 1250. So e² = 1 + 1250/25 = 1 + 50.
4. So. The square of the eccentricity is 51.

The common slip: 50. Reporting B²/A² = a² and leaving out the 1 in e² = 1 + B²/A² gives 50, not 51.

The idea to keep. For x²/A² - y²/B² = 1, use e² = 1 + B²/A². Always simplify that ratio before you put in numbers, because it often cancels down to something easy.

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