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 (2a, 0) 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

We know from the chapter ellipse, the equation of a tangent of slope “m” at any point is given as: $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$. It is given that this equation of tangent is passing through point (2a, 0) so substituting the value of y as 0 and x as 2a in this equation of tangent. We will also require the formula for eccentricity of the ellipse which we are denoting by “e” and the relation is as follows: ${{e}^{2}}=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$. And then rearrange this equation to get the required relation.

Complete step by step answer:
The equation of ellipse given in the above problem is as follows:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
And tangent of a slope “m” is given at a point on the ellipse. We know that the equation of tangent at any point with slope “m” is given in the chapter ellipse is as follows:
$y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
It is also given that this tangent equation is passing through point $\left( 2a,0 \right)$ so passing this point in the above equation by substituting the value of x as $2a$ and y as 0 in the above equation and we get,
$\Rightarrow 0=m\left( 2a \right)\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
Subtracting $2ma$ on both the sides of the above equation and we get,
$-2ma=\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
Squaring on both the sides of the above equation we get,
$\begin{aligned} & {{\left( -2ma \right)}^{2}}={{\left( \pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}} \right)}^{2}} \\\ & \Rightarrow 4{{m}^{2}}{{a}^{2}}=\left( {{a}^{2}}{{m}^{2}}+{{b}^{2}} \right)...........(1) \\\ \end{aligned}$
We know the eccentricity of the ellipse as:
${{e}^{2}}=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$
Rearranging the above equation we get,
$\dfrac{{{b}^{2}}}{{{a}^{2}}}=1-{{e}^{2}}$ …………. (2)
Now, rearranging the eq. (1) we get,
$4{{m}^{2}}{{a}^{2}}=\left( {{a}^{2}}{{m}^{2}}+{{b}^{2}} \right)$
Dividing ${{a}^{2}}$ on both the sides of the above equation we get,
$\begin{aligned} & \dfrac{4{{m}^{2}}{{a}^{2}}}{{{a}^{2}}}=\left( \dfrac{{{a}^{2}}{{m}^{2}}}{{{a}^{2}}}+\dfrac{{{b}^{2}}}{{{a}^{2}}} \right) \\\ & \Rightarrow 4{{m}^{2}}={{m}^{2}}+\dfrac{{{b}^{2}}}{{{a}^{2}}} \\\ \end{aligned}$
Using eq. (2) in the above equation we get,
$\begin{aligned} & 4{{m}^{2}}={{m}^{2}}+1-{{e}^{2}} \\\ & \Rightarrow 3{{m}^{2}}+{{e}^{2}}=1 \\\ \end{aligned}$

So, the correct answer is “Option c”.

Note: To solve the above problem, you need to know the equation of a tangent of slope m at any point of the ellipse. Secondly, you should also know the formula for eccentricity of the ellipse. Failing any of the two formulas will not give you desired results.