IIT-JEE 2012 · Paper 1 · Mathematics Q46

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MathematicsSingle correct+3 / -1Question 46 of 60

The ellipse $E_1:\dfrac{x^2}9+\dfrac{y^2}4=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through the point $(0,4)$ circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_2$ is

\dfrac{\sqrt2}2

\dfrac{\sqrt3}2

\dfrac12

\dfrac34

Answer and worked solution

Official answer: C

Rectangle. $E_1$ has semi-axes 3 and 2, so $R$ has vertices $(\pm3,\pm2)$ .

$E_2$ . Take $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ . Through $(0,4)$ : $B=4$ . Through $(3,2)$ : $\frac9{A^2}+\frac4{16}=1$ , so $A^2=12$ .

Eccentricity. The major axis is vertical ( $B>A$ ):
$e=\sqrt{1-\frac{A^2}{B^2}}=\sqrt{1-\frac{12}{16}}=\frac12.$

(C)

Topics:Ellipse

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IIT-JEE 2012 · Paper 1, Mathematics question 46