Question

The length of sub tangent corresponding to the point $\left( {3,\dfrac{{12}}{5}} \right)$ on the ellipse is $\dfrac{{16}}{3}$. Then the eccentricity of the ellipse is:
(A) $\dfrac{4}{5}$
(B) $\dfrac{2}{3}$
(C) $\dfrac{1}{5}$
(D) $\dfrac{3}{5}$

Answer 1

Hint : In the given question, we are required to find the eccentricity of the ellipse provided that the length of sub tangent corresponding to the point $\left( {3,\dfrac{{12}}{5}} \right)$ on the ellipse is $\dfrac{{16}}{3}$. To find the eccentricity of the ellipse, we need to find the length of subtangent in terms a and b of the ellipse and then equate it with the given length of subtangent and simplify to get the values of a and b so that we can get the equation of ellipse such that we can find the value of eccentricity of the ellipse.

Complete step by step solution:
Let ellipse equation $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.
Now, we differentiate the equation of ellipse given above with respect to x and find the value of $\dfrac{{dy}}{{dx}}$ in terms of a and b.
So, we get,
$\dfrac{{2x}}{{{a^2}}} + \dfrac{{2y}}{{{b^2}}}\dfrac{{dy}}{{dx}} = 0$
$\Rightarrow \dfrac{{2y}}{{{b^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{{a^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - \left( {\dfrac{{{b^2}x}}{{y{a^2}}}} \right)$
$\Rightarrow \dfrac{{dx}}{{dy}} = - \left( {\dfrac{{{a^2}y}}{{x{b^2}}}} \right)$

Length of sub tangent $= \left| {y\dfrac{{dx}}{{dy}}} \right| = \left| { - \dfrac{{{a^2}{y^2}}}{{x{b^2}}}} \right|$
Now, equating the above expression with the length of sub tangent given to us in the question,
$\Rightarrow \left| {\dfrac{{{a^2}{{\left( {\dfrac{{12}}{5}} \right)}^2}}}{{3{b^2}}}} \right| = \dfrac{{16}}{3}$
$\Rightarrow \dfrac{{144{a^2}}}{{75{b^2}}} = \dfrac{{16}}{3}$
$\Rightarrow \dfrac{{{a^2}}}{{{b^2}}} = \dfrac{{25}}{9}$
$\Rightarrow \left( {\dfrac{a}{b}} \right) = \sqrt {\dfrac{{25}}{9}} = \left( {\dfrac{5}{3}} \right)$
So, finding the eccentricity of the ellipse, we get,
$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}}$
$\Rightarrow e = \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}} = \sqrt {1 - \dfrac{9}{{25}}}$
$\Rightarrow e = \sqrt {\dfrac{{16}}{{25}}}$
$\Rightarrow e = \dfrac{4}{5}$
Hence the value of eccentricity is $\left( {\dfrac{4}{5}} \right)$ .
Therefore option A is the correct answer for this problem.
So, the correct answer is “Option A”.

Note : In this problem we cannot use the direct formula of eccentricity because the value of major axis and minor axis length is not given. So due to this reason, we have to start solving by length of subtangent concepts and reach the formula of eccentricity.