Let the product of the focal distances of the point (√3, 1/2) on the ellipse x^2/a^2 + y^2/b^2=1, (a>b), be 7/4.

Question

Let the product of the focal distances of the point \(\left(\sqrt{3}, \frac{1}{2}\right)\) on the ellipse \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1,(\mathrm{a}>\mathrm{b}), be \frac{7}{4}.\)

Then the absolute difference of the eccentricities of two such ellipses is

(1) \(\frac{3-2 \sqrt{2}}{3 \sqrt{2}}\)

(2) \(\frac{1-\sqrt{3}}{\sqrt{2}}\)

(3) \(\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\)

(4) \(\frac{1-2 \sqrt{2}}{\sqrt{3}}\)

Answer 1

Correct option is (3) \(\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\)   

\(\text{Given} (a+e \sqrt{3})(a-e \sqrt{3})=\frac{7}{4} \)  ...(i)

Also, \(\frac{3}{a^2}+\frac{1}{4 b^2}=1\)   ....(ii)

And \(b^2=a^2\left(1-e^2\right)\)   ....(iii)

From (i), (ii) and (iii)

\(12 e^4-17 e^2+6=0 \)

\(\Rightarrow\left(3 e^2-2\right)\left(4 e^2-3\right)=0 \Rightarrow e=\sqrt{\frac{2}{3}} \text { or } \sqrt{\frac{3}{4}}\)  

 \(\therefore\) difference \(=\frac{\sqrt3}{2}-\sqrt{\frac{2}{3}}=\frac{3-2\sqrt2}{2\sqrt3}\)