Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$
So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$
2
AIEEE 2010
MCQ (Single Correct Answer)
If two tangents drawn from a point $$P$$ to the parabola $${y^2} = 4x$$ are at right angles, then the locus of $$P$$ is
A
$$2x+1=0$$
B
$$x=-1$$
C
$$2x-1=0$$
D
$$x=1$$
Explanation
The locus of perpendicular tangents is directrix
i.e., $$x=-1$$
3
AIEEE 2009
MCQ (Single Correct Answer)
The ellipse $${x^2} + 4{y^2} = 4$$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $$(4,0)$$. Then the equation of the ellipse is :
A
$${x^2} + 12{y^2} = 16$$
B
$$4{x^2} + 48{y^2} = 48$$
C
$$4{x^2} + 64{y^2} = 48$$
D
$${x^2} + 16{y^2} = 16$$
Explanation
The given ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$
So $$A=(2,0)$$ and $$B = \left( {0,1} \right)$$
If $$PQRS$$ is the rectangular in which it is inscribed, then
$$P = \left( {2,1} \right).$$
Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
be the ellipse circumscribing the rectangular $$PQRS$$.