JEE Main 2017 · 9 Apr · Shift 1 · Mathematics Q80

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MathematicsSingle correct+4 / -1Question 80 of 90

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points (4, −1) and (−2, 2) is :

\dfrac12

\dfrac2{\sqrt5}

\dfrac{\sqrt3}2

\dfrac{\sqrt3}4

Answer and worked solution

Official answer: C

With $p=\frac1{a^2}$ , $q=\frac1{b^2}$ : $16p+q=1$ and $4p+4q=1$ , so $p=\frac1{20}$ , $q=\frac15$ . Then $a^2=20$ , $b^2=5$ and $e=\sqrt{1-\frac5{20}}=\frac{\sqrt3}2$ .

Topics:Ellipse

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JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 80