# If a Latus Rectum of an Ellipse Subtends a Right Angle at the Centre of the Ellipse, Then Write the Eccentricity of the Ellipse. - Mathematics

If a latus rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of the ellipse.

#### Solution

$\text{ According to the Pythogoras theorem, we have }:$
$O A^2 + O B^2 = A B^2$
$\text{ From the figure, we can see that }$
$OA = \sqrt{\left( \frac{b^2}{a} - 0 \right)^2 + \left( ae - 0 \right)^2} = \sqrt{\frac{b^4}{a^2}} + a^2 e^2 = OB and AB = \frac{2 b^2}{a}$
$\text{ Now }, 2\left[ a^2 e^2 + \frac{b^4}{a^2} \right] = \frac{4 b^4}{a^2}$
$\Rightarrow a^2 e^2 + \frac{b^4}{a^2} = \frac{2 b^4}{a^2}$
$\Rightarrow a^2 e^2 = - \frac{b^4}{a^2} + \frac{2 b^4}{a^2}$
$\Rightarrow a^2 e^2 = \frac{b^4}{a^2}$
$\Rightarrow e^2 = \frac{b^4}{a^4}$
$\Rightarrow e = \frac{b^2}{a^2}$
$\text{ We know that } e = \sqrt{1 - \frac{b^2}{a^2}}$
$e = \sqrt{1 - e}$
$\text{ On squaring both sides, we get }:$
$e^2 + e - 1 = 0$
$\Rightarrow e = \frac{- 1 \pm \sqrt{1 + 4}}{2} \left( \because \text{ Ecentricity cannot be negative } \right)$
$\Rightarrow e = \frac{\sqrt{5} - 1}{2}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Q 9 | Page 27