If the foci of an ellipse are (pm sqrt{5}, 0) | vedclass

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

Question

(B) Given the foci are $(\pm \sqrt{5}, 0)$,we have $ae = \sqrt{5}$.Given eccentricity $e = \frac{\sqrt{5}}{3}$.Substituting $e$ in $ae = \sqrt{5}$,we

Vedclass · Accepted Answer

$4x^2 + 9y^2 = 36$

Vedclass · Answer

(B) Given the foci are $(\pm \sqrt{5}, 0)$,we have $ae = \sqrt{5}$.Given eccentricity $e = \frac{\sqrt{5}}{3}$.Substituting $e$ in $ae = \sqrt{5}$,we get $a(\frac{\sqrt{5}}{3}) = \sqrt{5}$,which implies $a = 3$,so $a^2 = 9$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 9(1 - \frac{5}{9}) = 9(\frac{4}{9}) = 4$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which becomes $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Multiplying by $36$,we get $4x^2 + 9y^2 = 36$.

If the foci of an ellipse are $(\pm \sqrt{5}, 0)$ and its eccentricity is $\frac{\sqrt{5}}{3}$,then the equation of the ellipse is

Similar Questions

If $(1, -2)$ is the focus and $x+y-2=0$ is the directrix of the ellipse $17x^2 - 2xy + 17y^2 - 32x + 76y + 86 = 0$,then its eccentricity is

If the length of the minor axis of an ellipse is equal to one-fourth of the distance between the foci,then the eccentricity of the ellipse is:

The equation of the tangent at the point $(1/4, 1/4)$ of the ellipse $\frac{x^2}{4} + \frac{y^2}{12} = 1$ is

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$.

If tangents are drawn from any point on the circle $x^2+y^2=25$ to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$,then the angle between the tangents is

Vedclass Test Series

Exam Paper Generator

Online Exam Module