MHT CET · Maths · Ellipse
If \(\mathrm{B}\) is end point of minor axis of the ellipse \(b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}(a>b)\) and \(\mathrm{S}\) and S' are foci of ellipse such that \(\Delta \mathrm{SBS}^{\prime}\) is an equilateral triangle, then eccentricity \(\mathrm{e}=\)
- A \(\frac{1}{2}\)
- B \(\frac{1}{3}\)
- C \(\frac{3}{5}\)
- D \(\frac{4}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Correct option is A
Given \(\mathrm{S}(-\mathrm{ae}, 0), \mathrm{T}(\mathrm{ae}, 0), \mathrm{B}(0, \mathrm{~b})\)
As STB is an equilateral triangle
In \(\triangle \mathrm{TOB}\)
\(\tan 60^{\circ}=\frac{\mathrm{OB}}{\mathrm{OS}}\)
\(\sqrt{3}=\frac{\mathrm{OB}}{\mathrm{OS}}=\frac{\mathrm{b}}{\mathrm{ae}}\)
\((\sqrt{3} a e)^{2}=b^{2}\)
\(3 a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=b^{2}\)
\(3\left(1-\frac{b^{2}}{a^{2}}\right)=\frac{b^{2}}{a^{2}}\)
\(\frac{3}{4}=\frac{b^{2}}{a^{2}}\)
\(e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\)
Given \(\mathrm{S}(-\mathrm{ae}, 0), \mathrm{T}(\mathrm{ae}, 0), \mathrm{B}(0, \mathrm{~b})\)
As STB is an equilateral triangle
In \(\triangle \mathrm{TOB}\)
\(\tan 60^{\circ}=\frac{\mathrm{OB}}{\mathrm{OS}}\)
\(\sqrt{3}=\frac{\mathrm{OB}}{\mathrm{OS}}=\frac{\mathrm{b}}{\mathrm{ae}}\)
\((\sqrt{3} a e)^{2}=b^{2}\)
\(3 a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=b^{2}\)
\(3\left(1-\frac{b^{2}}{a^{2}}\right)=\frac{b^{2}}{a^{2}}\)
\(\frac{3}{4}=\frac{b^{2}}{a^{2}}\)
\(e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\)
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