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JEE Main 2009

Exam: JEE Main
Year: 2009
Subject: Mathematics
Chapter: Ellipse

JEE Main 2009 · Mathematics · Ellipse

The ellipse x^2+4 y^2=4 is inscribed in a rectangle aligned with...

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 The ellipse x^2+4 y^2=4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in

The ellipse x^2+4 y^2=4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in

The ellipse x^2+4 y^2=4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in

Options

A $x^2+16 y^2=16$
B $x^2+12 y^2=16$
C $4 x^2+48 y^2=48$
D $4 \mathrm{x}^2+64 \mathrm{y}^2=48$

Correct answer

B. $x^2+12 y^2=16$

Step-by-step solution

x²+4 y²=4 \Rightarrow

\frac{x^2}{4}+\frac{y^2}{1}=

1 \Rightarrow a=2, b=1 \Rightarrow P=(2,1) Required Ellipse is

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{4^2}+\frac{y^2}{b^2}=1(2,1)

lies on it

\begin{aligned}& \Rightarrow \frac{4}{16}+\frac{1}{b^2}=1 \Rightarrow \frac{1}{b^2}=1-\frac{1}{4}=\frac{3}{4} \Rightarrowb^2=\frac{4}{3}\\ & \therefore \frac{x^2}{16}+\frac{y^2}{\left(\frac{4}{3}\right)}=1 \Rightarrow \frac{x^2}{16}+\frac{3 y^2}{4}=1 \Rightarrow x²+12 y²=16 \end{aligned}

The ellipse x^2+4 y^2=4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in

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