The eccentricity of the ellipse 9x^2 + 5y^2 - | vedclass

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

Question

(B) Given equation: $9x^2 + 5y^2 - 30y = 0$ Complete the square for $y$: $9x^2 + 5(y^2 - 6y) = 0$ $9x^2 + 5(y^2 - 6y + 9) = 45$ $9x^2 + 5(y - 3)^2 = 4

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(B) Given equation: $9x^2 + 5y^2 - 30y = 0$ Complete the square for $y$: $9x^2 + 5(y^2 - 6y) = 0$ $9x^2 + 5(y^2 - 6y + 9) = 45$ $9x^2 + 5(y - 3)^2 = 45$ Divide by $45$: $\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1$ Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the ellipse is vertical. The eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$ $5 = 9(1 - e^2)$ $1 - e^2 = \frac{5}{9}$ $e^2 = 1 - \frac{5}{9} = \frac{4}{9}$ $e = \sqrt{\frac{4}{9}} = \frac{2}{3}$

The eccentricity of the ellipse $9x^2 + 5y^2 - 30y = 0$ is ....

Similar Questions

If the length of the latus rectum of an ellipse is $4 \ units$ and the distance between a focus and its nearest vertex on the major axis is $\frac{3}{2} \ units$,then its eccentricity is?

The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

If the lines $2x - y + 3 = 0$ and $4x + ky + 3 = 0$ are conjugate with respect to the ellipse $5x^2 + 6y^2 - 15 = 0$,then $k$ equals:

The ellipse $4x^2 + 9y^2 = 1$ intersects the positive $Y$-axis at point $A$. If $S$ and $S'$ are its foci, then the area (in sq. units) of $\Delta SAS'$ is

Find the coordinates of the foci,the vertices,the lengths of the major and minor axes,and the eccentricity of the ellipse $9x^{2} + 4y^{2} = 36$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module