Q96 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics
If the latusrectum of an ellipse subtends a right angle at the centre of that ellipse, then the eccentricity of that ellipse is
- A$\frac{\sqrt{5}+1}{4}$
- Correct answerB$\frac{\sqrt{5}-1}{2}$
- C$\frac{\sqrt{10-2 \sqrt{5}}}{5}$
- D$\frac{\sqrt{10+2 \sqrt{5}}}{5}$
Solution
Let equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $L L^{\prime}$ is latusrectum, then co-ordinate of $$ L\left(a e, \frac{b^2}{a}\right) $$ $L L^{\prime}$ subtend $\pi / 2$ Angle at the centre, so angle $$ L C S=\pi / 4 $$ $$ \begin{array}{rlrl} \Rightarrow & \tan \frac{\pi}{4} & =\frac{\frac{b^2}{a}}{a e} \\ \Rightarrow & 1 & =\frac{b^2}{a^2 e} \\ \Rightarrow & a^2 e & =b^2=a^2\left(1-e^2\right) \\ \Rightarrow & e & =1-e^2 \\ \Rightarrow & e^2+e-1 & =0 \\ \Rightarrow & & e & =\frac{-1 \pm \sqrt{5}}{2} \Rightarrow e=\frac{\sqrt{5}-1}{2} . \end{array} $$
Let $L L^{\prime}$ is latusrectum, then co-ordinate of $$ L\left(a e, \frac{b^2}{a}\right) $$ $L L^{\prime}$ subtend $\pi / 2$ Angle at the centre, so angle $$ L C S=\pi / 4 $$ $$ \begin{array}{rlrl} \Rightarrow & \tan \frac{\pi}{4} & =\frac{\frac{b^2}{a}}{a e} \\ \Rightarrow & 1 & =\frac{b^2}{a^2 e} \\ \Rightarrow & a^2 e & =b^2=a^2\left(1-e^2\right) \\ \Rightarrow & e & =1-e^2 \\ \Rightarrow & e^2+e-1 & =0 \\ \Rightarrow & & e & =\frac{-1 \pm \sqrt{5}}{2} \Rightarrow e=\frac{\sqrt{5}-1}{2} . \end{array} $$
Hence, the correct answer is option: 'B'.