Question

If a tangent of slope $m$ at a point of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ passes through $\left( 2a,0 \right)$ and if “$e$” denotes the eccentricity of the ellipse then
(A) ${{m}^{2}}+{{e}^{2}}=1$
(B) $2{{m}^{2}}+{{e}^{2}}=1$
(C) $3{{m}^{2}}+{{e}^{2}}=1$
(D) ${{m}^{2}}+{{e}^{2}}-2=0$

Answer 1

Hint: For answering this question we will use the formulae for the equation of any tangent to the given ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ with slope $m$ is given by $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$ . From the basic concepts we can say that the eccentricity of the ellipse is given by the formulae $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ . By using them we will answer this question.

Complete step-by-step answer:
Now considering from the question we have a tangent of slope $m$ at a point of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ passes through $\left( 2a,0 \right)$ and “$e$” denotes the eccentricity of the ellipse.
From the basics concept we can say that the equation of any tangent to the given ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ with slope $m$ is given by $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$

Since tangent with slope $m$ passes through the point $\left( 2a,0 \right)$, then the equation of the tangent to the ellipse is given by substituting the values in the equation $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$ we will have
 $0=2am\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
By shifting it to left hand side and squaring on both sides we will have
$4{{a}^{2}}{{m}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}$ ….$\left( 1 \right)$
From the basic concepts we can say that the eccentricity of the ellipse is given by $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ .
From this formula of eccentricity we can get,
$\begin{align}
  & {{e}^{2}}=1-\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
 & \Rightarrow {{b}^{2}}={{a}^{2}}\left( 1-{{e}^{2}} \right) \\
\end{align}$
Now, by substituting this value of ${{b}^{2}}$ in …….$\left( 1 \right)$
We will get
  $\begin{align}
  & 4{{a}^{2}}{{m}^{2}}={{a}^{2}}{{m}^{2}}+{{a}^{2}}\left( 1-{{e}^{2}} \right) \\
 & \Rightarrow 3{{a}^{2}}{{m}^{2}}={{a}^{2}}\left( 1-{{e}^{2}} \right) \\
 & \Rightarrow 2{{m}^{2}}=1-{{e}^{2}} \\
 & \Rightarrow 2{{m}^{2}}+{{e}^{2}}=1 \\
\end{align}$
Hence, option C is the correct option.

So, the correct answer is “Option C”.

Note: While answering this type of questions of this type we should be sure with the calculations and formulae. From the basic concepts we can say that the eccentricity of the ellipse is given by $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ . The eccentricity for different curves is different like for the hyperbola it is given as $e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ and similarly for parabola it is given as $1$ .