Let P and Q be the foci of an ellipse and let | vedclass

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

Question

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The foci are $P = (-ae, 0)$ and $Q = (ae, 0)$.The end of

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The foci are $P = (-ae, 0)$ and $Q = (ae, 0)$.The end of the minor axis is $R = (0, b)$.Since $	riangle PQR$ is an equilateral triangle,the distance $PQ = PR = QR$.The distance $PQ = 2ae$.The distance $PR = \sqrt{(ae - 0)^2 + (0 - b)^2} = \sqrt{a^2e^2 + b^2}$.Equating $PQ^2 = PR^2$,we get $(2ae)^2 = a^2e^2 + b^2$.$4a^2e^2 = a^2e^2 + b^2 \implies 3a^2e^2 = b^2$.Using the relation $b^2 = a^2(1 - e^2)$,we have $3a^2e^2 = a^2(1 - e^2)$.$3e^2 = 1 - e^2 \implies 4e^2 = 1 \implies e^2 = \frac{1}{4}$.Thus,$e = \frac{1}{2}$.

Let $P$ and $Q$ be the foci of an ellipse and let $R$ be one end of its minor axis. If $\triangle PQR$ is an equilateral triangle,then the eccentricity of the ellipse is equal to

Similar Questions

For $k>0$,the shortest distance from a point $P(1, k)$ on the ellipse $9x^2+4y^2-18x+16y-11=0$ to one of its directrices is

The eccentric angle of the point $(2, \sqrt{3})$ lying on $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ is

The equations of the tangents to the ellipse $9x^2 + 16y^2 = 144$ which pass through the point $(2, 3)$ are

The length of the latus rectum of the ellipse $4x^2 + 9y^2 - 8x - 36y + 4 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module