Question

If the length of minor axis (along y-axis) of an ellipse of the standard form is $\dfrac{4}{{\sqrt 3 }}$.If this ellipse touches the line $x + 6y = 8$,then what is its eccentricity?
A. $\dfrac{1}{2}\left( {\sqrt {\dfrac{5}{3}} } \right)$
B. $\dfrac{1}{2}\left( {\sqrt {\dfrac{{11}}{3}} } \right)$
C. $\dfrac{{\sqrt 5 }}{6}$
D. $\dfrac{1}{3}\sqrt {\left( {\dfrac{{11}}{6}} \right)} $

Answer 1

Hint: We know the length of the minor axis of ellipse so from this we can get the value of $b$. Also, the equation of tangent is given to the ellipse, we can get the equation of tangent at any point $\left( {{x_1},{y_1}} \right)$ on ellipse by $\dfrac{{x.{x_1}}}{{{a^2}}} + \dfrac{{y.{y_1}}}{{{b^2}}} = 1$. We already know the value of $b$. On comparing the coefficient of equations of tangent to the given equation and solving them ,we can get the value of $a$.

Formula Used:
1.Eccentricity of ellipse is given by
 ${e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}}$
Where,
$e$ is eccentricity of ellipse
$a$ is semi major axis length of ellipse
 $b$is the length of semi minor axis of ellipse
2.Equation to the tangent on ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ at any point $\left( {{x_1},{y_1}} \right)$is given by –
$\dfrac{{x.{x_1}}}{{{a^2}}} + \dfrac{{y.{y_1}}}{{{b^2}}} = 1$

Complete step by step solution:
Given – Length of major axis =$\dfrac{4}{{\sqrt 3 }}$
$2b = \dfrac{4}{{\sqrt 3 }}$
$b = \dfrac{2}{{\sqrt 3 }}$
Now equation of tangent at a point $\left( {{x_1},{y_1}} \right)$is given can be written as
$\dfrac{{x.{x_1}}}{{{a^2}}} + \dfrac{{y.{y_1}}}{{{b^2}}} = 1$
Substituting the value of $b = \dfrac{2}{{\sqrt 3 }}$ in the above equation
$\dfrac{{x.{x_1}}}{{{a^2}}} + \dfrac{{y.{y_1}}}{{{{\left( {\dfrac{2}{{\sqrt 3 }}} \right)}^2}}} = 1$
$ \Rightarrow \dfrac{{x.{x_1}}}{{{a^2}}} + \dfrac{{3y.{y_1}}}{4} = 1$ (1)
Given equation of tangent is –
$x + 6y = 8$
$ \Rightarrow \dfrac{x}{8} + \dfrac{3}{4}y = 1$ (2)
Equation (1) and (2) represents same equations so on comparing them
$\dfrac{{{x_1}}}{{{a^2}}} = \dfrac{1}{8}$
On cross multiplying the equation –
${x_1} = \dfrac{{{a^2}}}{8}$ (3)
Also, by comparing the coefficients of Y in equation (1) and (2)
$\dfrac{{3{y_1}}}{4} = \dfrac{3}{4}$
$ \Rightarrow {y_1} = 1$
Point $\left( {{x_1},{y_1}} \right)$lies on ellipse so –
$\dfrac{{{{\left( {\dfrac{{{a^2}}}{8}} \right)}^2}}}{{{a^2}}}\,\,\,\,\, + \,\,\,\dfrac{{{{\left( 1 \right)}^2}}}{{{{\left( {\dfrac{2}{{\sqrt 3 }}} \right)}^2}}} = 1$
$ \Rightarrow \dfrac{{{a^2}}}{{64}} + \dfrac{3}{4} = 1$
$ \Rightarrow \dfrac{{{a^2}}}{{64}} = \dfrac{1}{4}$
$ \Rightarrow {a^2} = \dfrac{{64}}{4}$
$ \Rightarrow \,{a^2} = 16$
Now eccentricity of ellipse is given by
$e = 1 - \dfrac{{{b^2}}}{{{a^2}}}$
Putting ${a^2} = 16$ and ${b^2} = \dfrac{4}{3}$ in the above formula
${e^2} = 1 - \dfrac{{\dfrac{4}{3}}}{{16}}$
$ \Rightarrow {e^2} = 1 - \dfrac{1}{{12}}$
$ \Rightarrow {e^2} = \dfrac{{11}}{{12}}$
$ \Rightarrow e = \sqrt {\dfrac{{11}}{{12}}} $
$ \Rightarrow e = \dfrac{1}{2}\sqrt {\left( {\dfrac{{11}}{3}} \right)} $

Option ‘B’ is correct

Note: Students get confusion in coefficient comparison of equations. Let ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ be two equations representing same line then:
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$