%%

The length of the minor axis (along y-axis) of an ellipse in the standard form is $${4 \over {\sqrt 3 }}$$. If this ellipse touches the line, x + 6y = 8; then its eccentricity is :

JEE · Math · previous-year question

  1. A.$${1 \over 3}\sqrt {{{11} \over 3}} $$
  2. B.$${1 \over 2}\sqrt {{5 \over 3}} $$
  3. C.$$\sqrt {{5 \over 6}} $$
  4. D.$${1 \over 2}\sqrt {{{11} \over 3}} $$correct

Answer

D. $${1 \over 2}\sqrt {{{11} \over 3}} $$

Explanation

Let the equation of ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, ($$a > b$$) Given 2b = $${4 \over {\sqrt 3 }}$$ $$ \Rightarrow $$ b = $${2 \over {\sqrt 3 }}$$ We know, Equation of tangent y = mx $$ \pm $$ $$\sqrt {{a^2}{m^2} + {b^2}} $$ ....(1) Given tangent is x + 6y = 8 $$ \Rightarrow $$ y = $$ - {1 \over 6}x + {8 \over 6}$$ .....(2) By comparing (1) and (2), m = $$ - {1 \over 6}$$ and $${{a^2}{m^2} + {b^2}}$$ = $${{16} \over 9}$$ $$ \Rightarrow $$ $${{a^2}\left( {{1 \over {36}}} \right) + {4 \over 3}}$$ = $${{16} \over 9}$$ $$ \Rightarrow $$ $${{a^2} = 16}$$ $$ \therefore $$ e = $$\sqrt {1 - {{{b^2}} \over {{a^2}}}} $$ = $$\sqrt {1 - {{{4 \over 3}} \over {16}}} $$ = $$\sqrt {{{11} \over {12}}} $$ = $${1 \over 2}\sqrt {{{11} \over 3}} $$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions