Let frac{x^2}{a^2}+frac{y^2}{b^2}=1, ab be a | vedclass

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

Question

(D) Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$.Since $	riangle ABC$ is equilateral,the angle $\angle BAC = 60^{\circ}$.Due to t

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$.Since $	riangle ABC$ is equilateral,the angle $\angle BAC = 60^{\circ}$.Due to the symmetry of the ellipse,the line $AO$ (the $y$-axis) bisects $\angle BAC$,so $\angle OAF_2 = 30^{\circ}$.In the right-angled $	riangle AOF_2$,we have $	an(30^{\circ}) = \frac{OF_2}{OA}$.Here,$OF_2 = ae$ and $OA = b$.So,$\frac{1}{\sqrt{3}} = \frac{ae}{b}$,which implies $b = \sqrt{3}ae$.Squaring both sides,$b^2 = 3a^2e^2$.Using the relation $b^2 = a^2(1-e^2)$,we get $a^2(1-e^2) = 3a^2e^2$.Dividing by $a^2$,we have $1-e^2 = 3e^2$,which simplifies to $4e^2 = 1$.Thus,$e^2 = \frac{1}{4}$,and since $e>0$,$e = \frac{1}{2}$.

Similar Questions

If the area of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{\lambda^{2}}=1$ is $20 \pi$ sq units,then $\lambda$ is

The eccentricity of the ellipse $\frac{(x - 1)^2}{9} + \frac{(y + 1)^2}{25} = 1$ is

The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9x^2 + 16y^2 = 144$ is

Let $\frac{x^2}{f(a^2 + 7a + 3)} + \frac{y^2}{f(3a + 15)} = 1$ represent an ellipse with major axis along the y-axis,where $f$ is a strictly decreasing positive function on $R$. If the set of all possible values of $a$ is $R - [\alpha, \beta]$,then $\alpha^2 + \beta^2$ is equal to:

Let the equation of an ellipse be $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$. Then,the radius of the circle with center $(0, \sqrt{2})$ and passing through the foci of the ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module