If the end points of the latus rectum of an e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The coordinates of the end points of the latus rectum are $(ae, \frac{b^2

Vedclass · Accepted Answer

$\frac{\sqrt{5} - 1}{2}$

Vedclass · Answer

(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The coordinates of the end points of the latus rectum are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$.Since these points are vertices of a square,the distance between them must be equal to the distance between the two end points of the latus rectum on the same side of the major axis,which is $2ae$.Thus,$2 	imes \frac{b^2}{a} = 2ae$.$\frac{b^2}{a} = ae$.Using $b^2 = a^2(1 - e^2)$,we get $\frac{a^2(1 - e^2)}{a} = ae$.$a(1 - e^2) = ae \Rightarrow 1 - e^2 = e$.$e^2 + e - 1 = 0$.Solving for $e$ using the quadratic formula,$e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since $e > 0$,we have $e = \frac{\sqrt{5} - 1}{2}$.

If the end points of the latus rectum of an ellipse are the vertices of a square,then the eccentricity of the ellipse will be -

Similar Questions

If the distance between the directrices of an ellipse is three times the distance between its foci,then the eccentricity of the ellipse is:

$A$ man running round a race-course notes that the sum of the distances of two flag-posts from him is always $10 \ m$ and the distance between the flag-posts is $8 \ m$. The area of the path he encloses in square metres is (in $\pi$)

Let $F_1$ and $F_2$ be the foci of an ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$. $A$ ray from $F_1$ strikes the elliptical mirror at point $P$ and is reflected. What is the equation of the angle bisector of the angle between the incident ray and the reflected ray?

The equation of the ellipse whose distance between the foci is $8$ and the distance between the directrices is $18$,is

The values of $c$ such that the line $y=4x+c$ touches the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module