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Question

Mathematics Question on Circle

SS and TT are the foci of an ellipse and BB is an end of the minor axis. If STBSTB is an equilateral triangle, then the eccentricity of the ellipse is

A

14\frac{1}{4}

B

13\frac{1}{3}

C

12\frac{1}{2}

D

23\frac{2}{3}

Answer

12\frac{1}{2}

Explanation

Solution

Let e of ellipse be x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,
SS is (ae,0),T(-a e, 0), T is (ae,0)(a e, 0) and BB is (0,b)(0, b).
SB=(0+ae)2+b2\Rightarrow S B=\sqrt{(0+a e)^{2}+b^{2}}
Also SB2=ST2S B^{2}=S T^{2}
4a2e2=a2e2+b2\Rightarrow 4 a^{2} e^{2}=a^{2} e^{2}+b^{2}
3a2e2=a2(1e2)\Rightarrow 3 a^{2} e^{2}=a^{2}\left(1-e^{2}\right)
=a2a2e2=a^{2}-a^{2} e^{2}
4a2e2=a2\Rightarrow 4 a^{2} e^{2}=a^{2}
e2=14\Rightarrow e^{2}=\frac{1}{4}
e=12\Rightarrow e=\frac{1}{2}