JEE Main2009MathematicsEllipseActual
The ellipse x^2+4 y^2=4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point (4,0) . Then the equation of the ellipse is
Options
- Ax^2+16 y^2=16
- Bx^2+12 y^2=16
- C4 x^2+48 y^2=48
- D4 x ^2+64 y ^2=48
Correct answer
B. x^2+12 y^2=16
Step-by-step solution
x^2+4 y^2=4 x^2 4 + y^2 1 =1 a=2, b=1 P=(2,1) Required Ellipse is x^2 a^2 + y^2 b^2 =1 x^2 4^2 + y^2 b^2 =1(2,1) lies on it aligned & 4 16 + 1 b^2 =1 1 b^2 =1- 1 4 = 3 4 b^2= 4 3 & x^2 16 + y^2 ( 4 3 ) =1 x^2 16 + 3 y^2 4 =1 x^2+12 y^2=16 aligned