If the area of the auxiliary circle of the el | vedclass

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

Question

(B) The area of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by $A_{e} = \pi ab$.The auxiliary circle of the ellipse is $x^{2}+y^{

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(B) The area of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by $A_{e} = \pi ab$.The auxiliary circle of the ellipse is $x^{2}+y^{2}=a^{2}$,and its area is $A_{c} = \pi a^{2}$.According to the problem,the area of the auxiliary circle is twice the area of the ellipse:$\pi a^{2} = 2(\pi ab)$$a^{2} = 2ab$Since $a 
eq 0$,we have $a = 2b$.The eccentricity $e$ of the ellipse is given by $e = \sqrt{1-\frac{b^{2}}{a^{2}}}$.Substituting $a = 2b$ into the formula:$e = \sqrt{1-\frac{b^{2}}{(2b)^{2}}} = \sqrt{1-\frac{b^{2}}{4b^{2}}} = \sqrt{1-\frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

If the area of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b)$ is twice the area of the ellipse,then the eccentricity of the ellipse is

Similar Questions

If a solid sphere and a solid cylinder of the same radius and density rotate about their own axes,which one will have a greater moment of inertia? (Assume $L=R$ for the cylinder)

The major product in the following reaction is

An incompressible liquid flows through a horizontal tube as shown in the figure. Then,the velocity $v$ of the fluid is ......... $m/s$.

$74.5 \ g$ of a metallic chloride contain $35.5 \ g$ of chlorine. The equivalent weight of the metal is

If the density of the earth is doubled keeping its radius constant,then acceleration due to gravity will be ........ $m/s^2$. $(g = 9.8\,m/s^2)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module