JEE MainMathematicsHyperbola
Let E: x^2 A^2 + y^2 B^2 = 1 be an ellipse passing through the point (0, 5) , and its chord parallel to the x-axis at y=3 has a length of 16 . The ellipse E passes through the foci of the hyperbola H: x^2 a^2 - y^2 b^2 = 1 . If the product of the eccentricities of E and H is 1 , then the length of the latus rectum of H is :
Options
- A5
- B2 15 3
- C30
- D10 3
Correct answer
D. 10 3
Step-by-step solution
Since the ellipse E: x^2 A^2 + y^2 B^2 = 1 passes through (0, 5) , its semi-minor axis is B = 5 . The chord parallel to the x-axis at y=3 has a length of 16 , which means the points ( 8, 3) lie on the ellipse. Substituting (8, 3) into the equation of E : 64 A^2 + 9 25 = 1 64 A^2 = 1 - 9 25 = 16 25 A^2 = 100 A = 10 . The eccentricity of the ellipse E is e₂ = 1 - B^2 A^2 = 1 - 25 100 = 3 2 . Given that the product of their eccentricities is 1 , the eccentricity of the hyperbola H is e₁ = 1 e₂ = 2 3 . The foci of the