QJEEM09Ma54b7: JEE Main 2009 Mathematics - Class 11 JEEM09Ma54b7 Solution: The ellipse is inscribed in a rectangle aligned with the…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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

.

Then it passes through

P\,\,(2,1)

\therefore

{4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)

Also, given that, it passes through

(4,0)

\therefore

{{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16

\Rightarrow {b^2} = 4/3

\left[ {\,\,} \right.

substituting

{{a^2} = 16\,\,}

in

\left. {e{q^n}\left( a \right)\,\,} \right]

\therefore

The required ellipse is

{{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1

or

{x^2} + 12y{}^2 = 16

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