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, $x+6 y=8$; then its eccentricity is:

(C) $\frac{1}{2} \sqrt{\frac{11}{3}}$

(A) $\frac{1}{3} \sqrt{\frac{11}{3}}$

(B) $\frac{1}{2} \sqrt{\frac{5}{3}}$

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Ellipse

The length of the minor axis (along

y

-axis) of an ellipse in the standard form is

\frac{4}{3}

. If this ellipse touches the line,

x + 6 y = 8

; then its eccentricity is:

Answer: (C)

The minor axis along

y

has length

\frac{4}{3}

, so

2 b = \frac{4}{3}

and

b^{2} = \frac{4}{3}

. Writing the line as

y = - \frac{1}{6} x + \frac{4}{3}

, its slope is

m = - \frac{1}{6}

and intercept

c = \frac{4}{3}

, and tangency to

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

requires

c^{2} = a^{2} m^{2} + b^{2}

. Then

\frac{16}{9} = \frac{a ^{2}}{36} + \frac{4}{3}

, giving

\frac{a ^{2}}{36} = \frac{4}{9}

and

a^{2} = 16

. The eccentricity is

e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{4/3}{16} = \frac{11}{12} = \frac{1}{2} \frac{11}{3}

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option