JEE PYQ: Ellipse Question 17
Question 17 - 2020 (09 Jan Shift 2)
The length of the minor axis (along y-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}$. If this ellipse touches the line, $x + 6y = 8$; then its eccentricity is:
(a) $\frac{1}{2}\sqrt{\frac{11}{3}}$ (b) $\sqrt{\frac{5}{6}}$ (c) $\frac{1}{2}\sqrt{\frac{5}{3}}$ (d) $\frac{1}{3}\sqrt{\frac{11}{3}}$
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Answer: (a) $\frac{1}{2}\sqrt{\frac{11}{3}}$
Solution
$2b = \frac{4}{\sqrt{3}} \Rightarrow b = \frac{2}{\sqrt{3}}$
Tangent: $y = \frac{-x}{6} + \frac{4}{3}$, so $m = \frac{-1}{6}$
$a^2m^2 + b^2 = \frac{16}{9} \Rightarrow \frac{a^2}{36} + \frac{4}{3} = \frac{16}{9} \Rightarrow a^2 = 16$
$e = \sqrt{1 - \frac{4}{48}} = \sqrt{\frac{11}{12}} = \frac{1}{2}\sqrt{\frac{11}{3}}$
BETA
