Question

Let $S$ and $S'$ be two foci of the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ . If a circle described on $SS'$ as diameter intersects the ellipse in real and distinct points, then the eccentricity e of the ellipse satisfies which of the following?
A) $e = \dfrac{1}{{\sqrt 2 }}$
B) $e \in \left( {\dfrac{1}{{\sqrt 2 }},1} \right)$
C) $e \in \left( {0,\dfrac{1}{{\sqrt 2 }}} \right)$
D) None of these

Answer 1

The foci of an ellipse with the equation $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ are $S\left( {ae,0} \right)$ and $S'\left( { - ae,0} \right)$ . Using these values find the equation of the circle with diameter as $SS'$ . Further, find the value of $x$ and $y$ . From these values, the value or limit of eccentricity can be calculated.

Complete step by step answer:
Given to us an ellipse with the equation $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
We know that the foci of this ellipse are $S\left( {ae,0} \right)$ and $S'\left( { - ae,0} \right)$
Now, we have been given that $SS'$ is the diameter of the circle that intersects the ellipse. We can find the equation for this circle by using the formula $\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$
So the circle would be $\left( {x - ae} \right)\left( {x + ae} \right) + \left( {y - 0} \right)\left( {y - 0} \right) = 0$
By solving, we get ${x^2} - {\left( {ae} \right)^2} + {y^2} = 0$
From this we get ${y^2} = {a^2}{e^2} - {x^2}$

Let us substitute this value in the equation of ellipse to find the value of $x$
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{a^2}{e^2} - {x^2}}}{{{b^2}}} = 1 \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{a^2}{e^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$
By taking ${x^2}$ common from this equation, we get ${x^2}\left( {\dfrac{1}{{{a^2}}} - \dfrac{1}{{{b^2}}}} \right) + \dfrac{{{a^2}{e^2}}}{{{b^2}}} = 1 \Rightarrow {x^2}\left( {\dfrac{1}{{{a^2}}} - \dfrac{1}{{{b^2}}}} \right) = 1 - \dfrac{{{a^2}{e^2}}}{{{b^2}}}$
We solve this as ${x^2}\left( {\dfrac{{{b^2} - {a^2}}}{{{a^2}{b^2}}}} \right) = \dfrac{{{b^2} - {a^2}{e^2}}}{{{b^2}}} \Rightarrow {x^2}\left( {\dfrac{{{b^2} - {a^2}}}{{{a^2}}}} \right) = \left( {{b^2} - {a^2}{e^2}} \right)$
This can be written as ${x^2}\left( {\dfrac{{{b^2}}}{{{a^2}}} - 1} \right) = {a^2}\left( {\dfrac{{{b^2}}}{{{a^2}}} - {e^2}} \right)$
We know that $\dfrac{b}{a} = \sqrt {1 - {e^2}} \Rightarrow \dfrac{{{b^2}}}{{{a^2}}} = 1 - {e^2}$

By substituting this value in the above equation, we get ${x^2}\left( {1 - {e^2} - 1} \right) = {a^2}\left( {1 - {e^2} - {e^2}} \right) \Rightarrow - {x^2}{e^2} = {a^2}\left( {1 - 2{e^2}} \right)$
Further solving this, ${x^2} = \dfrac{{{a^2}}}{{{e^2}}}\left( {2{e^2} - 1} \right)$
$x = \pm \dfrac{a}{e}\sqrt {2{e^2} - 1}$
For the value of $x$ to be real and distinct $2{e^2} - 1 > 0$
${e^2} > \dfrac{1}{2} \Rightarrow e > \pm \dfrac{1}{{\sqrt 2 }}$
But the value of $e$ lies between $\left( {0,1} \right)$
Therefore $e \in \left( {\dfrac{1}{{\sqrt 2 }},1} \right)$ i.e. option B.

Note: Note that the circle intersects the ellipse at real and distinct points so the values of $x$ and $y$ should be real and distinct. The limit of eccentricity for any ellipse is $\left( {0,1} \right)$ . The intersection of this limit and the limit calculated will be the resultant limit.