The eccentricity of the ellipse 4x^2 + 25y^2 | vedclass

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

Question

(A) Given equation of the ellipse is $4x^2 + 25y^2 = 100$. Dividing both sides by $100$,we get $\frac{4x^2}{100} + \frac{25y^2}{100} = \frac{100}{100}

Vedclass · Accepted Answer

$\frac{\sqrt{21}}{5}$

Vedclass · Answer

(A) Given equation of the ellipse is $4x^2 + 25y^2 = 100$. Dividing both sides by $100$,we get $\frac{4x^2}{100} + \frac{25y^2}{100} = \frac{100}{100}$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{4} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 4$. The eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting the values,$e = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{25 - 4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$. Thus,the correct option is $A$.

The eccentricity of the ellipse $4x^2 + 25y^2 = 100$ is

Similar Questions

The eccentricity of the conic $36x^2 + 144y^2 - 36x - 96y - 119 = 0$ is

The sum of focal radii of the curve $9x^{2} + 25y^{2} = 225$ is

If a tangent of slope $2$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ touches the circle $x^2+y^2=4$,then the maximum value of $ab$ is

The eccentricity of the conic $16x^2 + 7y^2 = 112$ is

If the points of intersection of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=4b$ (where $b > 4$) lie on the curve $y^{2}=3x^{2}$,then $b$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module