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

Find the coordinates of the foci of the ellipse $25(x + 1)^2 + 9(y + 2)^2 = 225$.

If the straight line $x \cos \alpha + y \sin \alpha = p$ touches the curve $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,then prove that $a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha = p^{2}$.

Find the equation for the ellipse that satisfies the given conditions: $b=3, c=4,$ centre at the origin; foci on the $x$-axis.

The acute angle between the pair of tangents drawn to the ellipse $2x^{2} + 3y^{2} = 5$ from the point $(1, 3)$ is:

$a$ and $b$ are the semi-major and semi-minor axes of an ellipse whose axes are along the coordinate axes. If its latus rectum is of length $4$ units and the distance between its foci is $4 \sqrt{2}$,then $a^2+b^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module