In an ellipse,if the distance between the foc | vedclass

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

Question

(D) Consider the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The distance between its foci is $2c = 6$,which implies $c

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

(D) Consider the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The distance between its foci is $2c = 6$,which implies $c = 3$. The length of the minor axis is $2b = 8$,which implies $b = 4$. For an ellipse,the relationship between $a, b,$ and $c$ is $a^2 = b^2 + c^2$. Substituting the values,we get $a^2 = 4^2 + 3^2 = 16 + 9 = 25$,so $a = 5$. The eccentricity $e$ is given by $e = \frac{c}{a}$. Therefore,$e = \frac{3}{5}$.

In an ellipse,if the distance between the foci is $6$ units and the length of its minor axis is $8$ units,then its eccentricity is

Similar Questions

The values that $m$ can take,so that the straight line $y=4x+m$ touches the curve $x^2+4y^2=4$ are

The locus of the point of intersection of perpendicular tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

If the point $P$ on the curve $4x^{2} + 5y^{2} = 20$ is farthest from the point $Q(0, -4)$,then $PQ^{2}$ is equal to:

The eccentricity of an ellipse,with its centre as origin,is $1/2$. If one of the directrices is $x=4$,then the equation of the ellipse is given by

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module