Question

. The ellipse x² + 4y2 = 4 is inscribed in a
rectangle aligned with the coordinate axes,
which is turn in inscribed in another ellipse that
passes through the point (4,0). Then, the
equation of the ellipse is
(EAM-2009)
1) x² +12 y2 = 16 2) 4x² + 48 y2 = 48
3) 4x² +64 y2 = 48 4) x2 +16y2 = 16​

Answer 1

Step-by-step explanation:

$\sf The \: given \: ellipse \: is {x}^{2} + 4{y}^{2} = 4 \\ \\ \sf \longrightarrow \frac{{x}^{2}}{4} + {y}^{2} = 1 \\ \\ \sf So \:a = 4 \: and \: b = 1 \\ \\ \sf So \: point\: A(0,1) \: and \: B(2,0) \\ \\ \sf Hence \: point \:C(2,1) \\ \\ \sf For \: outer \: ellipse \: one \: point \: D(4,0) \: is \: given \\ \sf which\:lie\:on\:the\:x-axis\:so\:it\:is\:one\:of\:the\:vertex \\ \\ \sf So \: a = 4 \\ \\ \sf So\: required \: equation \: of \: ellipse \: \\ \\ \sf \longrightarrow \frac{{x}^{2}}{16} + \frac{{y}^{2}}{{b}^{2}} = 1 \\ \\ \sf We\:know\:that\:point\:C\:lies\:on\:ellipse \\ \\ \sf \longrightarrow \frac{{2}^{2}}{16} + \frac{1}{{b}^{2}} = 1 \\ \\ \longrightarrow \frac{1}{4} + \frac{1}{{b}^{2}} = 1 \\ \\ \sf \longrightarrow \frac{1}{{b}^{2}} = 1 - \frac{1}{4} \\ \\ \sf \longrightarrow \frac{1}{{b}^{2}} = \frac{3}{4} \\ \\ \sf \longrightarrow {b}^{2} = \frac{4}{3} \\ \\ \sf So\:the\: equation\:of\: required\:ellipse \: is \\ \\ \sf \frac{{x}^{2}}{16}+ \frac{{y}^{2}}{4/3} = 1 \\ \\ \sf \longrightarrow \frac{{x}^{2}}{16} + \frac{3{y}^{2}}{4} = 1 \\ \\ \sf \longrightarrow {x}^{2} + 12{y}^{2} = 16 \\ \\ \sf So\: correct\:choice\:is\:option\:A$

Answer 2

Answer:

$$\begin{lgathered}\sf The \: given \: ellipse \: is {x}^{2} + 4{y}^{2} = 4 \\ \\ \sf \longrightarrow \frac{{x}^{2}}{4} + {y}^{2} = 1 \\ \\ \sf So \:a = 4 \: and \: b = 1 \\ \\ \sf So \: point\: A(0,1) \: and \: B(2,0) \\ \\ \sf Hence \: point \:C(2,1) \\ \\ \sf For \: outer \: ellipse \: one \: point \: D(4,0) \: is \: given \\ \sf which\:lie\:on\:the\:x-axis\:so\:it\:is\:one\:of\:the\:vertex \\ \\ \sf So \: a = 4 \\ \\ \sf So\: required \: equation \: of \: ellipse \: \\ \\ \sf \longrightarrow \frac{{x}^{2}}{16} + \frac{{y}^{2}}{{b}^{2}} = 1 \\ \\ \sf We\:know\:that\:point\:C\:lies\:on\:ellipse \\ \\ \sf \longrightarrow \frac{{2}^{2}}{16} + \frac{1}{{b}^{2}} = 1 \\ \\ \longrightarrow \frac{1}{4} + \frac{1}{{b}^{2}} = 1 \\ \\ \sf \longrightarrow \frac{1}{{b}^{2}} = 1 - \frac{1}{4} \\ \\ \sf \longrightarrow \frac{1}{{b}^{2}} = \frac{3}{4} \\ \\ \sf \longrightarrow {b}^{2} = \frac{4}{3} \\ \\ \sf So\:the\: equation\:of\: required\:ellipse \: is \\ \\ \sf \frac{{x}^{2}}{16}+ \frac{{y}^{2}}{4/3} = 1 \\ \\ \sf \longrightarrow \frac{{x}^{2}}{16} + \frac{3{y}^{2}}{4} = 1 \\ \\ \sf \longrightarrow {x}^{2} + 12{y}^{2} = 16 \\ \\ \sf So\: correct\:choice\:is\:option\:A\end