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) The given equation of the ellipse is $9x^2 + 5y^2 - 18x - 20y - 16 = 0$. Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 4y) = 16$. Completing

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) The given equation of the ellipse is $9x^2 + 5y^2 - 18x - 20y - 16 = 0$. Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 4y) = 16$. Completing the square,we have $9(x^2 - 2x + 1) + 5(y^2 - 4y + 4) = 16 + 9 + 20$. This simplifies to $9(x - 1)^2 + 5(y - 2)^2 = 45$. Dividing by $45$,we get $\frac{(x - 1)^2}{5} + \frac{(y - 2)^2}{9} = 1$. Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the ellipse is vertical. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

The eccentricity of the ellipse $9x^2 + 5y^2 - 18x - 20y - 16 = 0$ is:

Similar Questions

The equation $x^2 - 2\sqrt{3}xy + 2y^2 = 0$ represents two sides of a triangle. If the angle between them is $\frac{\pi}{3}$,then the perimeter of that triangle is

If $\frac{dy}{dx} = \frac{xy + y}{xy + x}$,then the solution of the differential equation is

The sensation of fatigue in the muscles after prolonged strenuous physical work is caused by

In plants,which elements are most readily mobilized?

Which of the following structures or regions is incorrectly paired with its function?

Vedclass Test Series

Exam Paper Generator

Online Exam Module