JEE MainMathematicsHyperbola
Let a hyperbola H be confocal with the ellipse E: x^2 25 + y^2 16 = 1 . If the hyperbola H passes through the point P (3, 5 2 ) , then the area of the triangle formed by the tangent to H at P and the coordinate axes is
Options
- A8 3
- B2 3
- C4 3
- D5 3
Correct answer
C. 4 3
Step-by-step solution
For the ellipse E , a^2 = 25 and b^2 = 16 . The square of the focal distance is c^2 = a^2 - b^2 = 25 - 16 = 9 . Since the hyperbola H is confocal with the ellipse, its foci are also at ( 3, 0) . The equation of the family of confocal conics can be written as x^2 a^2 - y^2 9 - a^2 = 1 , where a^2 The hyperbola passes through the point P (3, 5 2 ) . Substituting this into the equation gives: 9 a^2 - 25/4 9 - a^2 = 1 Let a^2 = t . Then: 9 t - 25 4(9 - t) = 1 36(9 - t) - 25t = 4t(9 - t) 324 - 36t - 25t = 36t - 4t^2 4t^