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Question: If the area of the auxiliary circle of the ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{...

If the area of the auxiliary circle of the ellipse x2a2+y2b2=1 (a >b)\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\ \left( a\ >b \right)is twice the area of ellipse, then the eccentricity of the ellipse is
a.13\dfrac{1}{\sqrt{3}}
b.12\dfrac{1}{2}
c.12\dfrac{1}{\sqrt{2}}
d.32\dfrac{\sqrt{3}}{2}

Explanation

Solution

Hint: Find the respective areas of auxiliary circle and that of ellipse and hence find relation of a and b, then from that find value of eccentricity using formula e = 1−b2a2e\ =\ \sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}.

Complete step-by-step answer:

In the question given an ellipse’s equation is given as x2a2+y2b2=1 \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\ where  a >b\ a\ >b, so we can say that it Is flattened at x-axis.
We are also given information that the area of the auxiliary circle is twice the area of the ellipse.
At first, we will learn about what is auxiliary circle. In simple terms, auxiliary circle is a circle whose center coincides with the center of the given ellipse and the circle i'll be described on the major axis of an ellipse as its diameter.
Here, the radius of the major axis is ‘a’ as in the question given that  a >b\ a\ >b. So, we can find the area of the circle πr2\pi {{r}^{2}} where r is the radius which is a. Hence, area is πa2\pi {{a}^{2}}.
For finding the area of ellipse we can do by using the formula which is product of π\pi ,
(major length2)×(minor length2)\left( \dfrac{\text{major length}}{2} \right)\times \left( \dfrac{\text{minor length}}{2} \right)
The major length of ellipse x2a2+y2b2=1 \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\ is 2a and minor length of ellipse is 2b.
So, the area of ellipse is,
π×(2a2)×(2b2) = πab\pi \times \left( \dfrac{2a}{2} \right)\times \left( \dfrac{2b}{2} \right)\ =\ \pi ab.
Now, we know the area of the ellipse is πab\pi ab and the area of the circle is πa2\pi {{a}^{2}}.
So, according to question,
πa2 = 2πab\pi {{a}^{2}}\ =\ 2\pi ab
Which can be written as a = 2ba\ =\ 2b
So, the faction ba = 12\dfrac{b}{a}\ =\ \dfrac{1}{2}.
In the question we have to find the eccentricity whose formula is,
e = 1−b2a2 = 1−(12)2e\ =\ \sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}\ =\ \sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}
So, the value of e = 34e\ =\ \sqrt{\dfrac{3}{4}} or 32\dfrac{\sqrt{3}}{2}.
Hence the correct option is (d).

Note: After finding the relationship between b and a, using the fact that the area of the auxiliary circle is twice that of the area of ellipse, in the expression of eccentricity instead of a put 2b.