The ellipse x2+4y2=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 :
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Official solution
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 x2+12y2=16