Let the equation of an ellipse be frac{x^{2}} | vedclass

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

Question

(C) The given equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$.Here, $a^{2}=144$ and $b^{2}=25$.The eccentricity $e$ is given by $e=\

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) The given equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$.Here, $a^{2}=144$ and $b^{2}=25$.The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}}=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}$.The foci of the ellipse are $(\pm ae, 0) = (\pm 12 	imes \frac{\sqrt{119}}{12}, 0) = (\pm \sqrt{119}, 0)$.The circle has its center at $(0, \sqrt{2})$ and passes through the foci $(\pm \sqrt{119}, 0)$.The radius $r$ of the circle is the distance between the center $(0, \sqrt{2})$ and one of the foci $(\sqrt{119}, 0)$.$r = \sqrt{(\sqrt{119}-0)^{2} + (0-\sqrt{2})^{2}}$$r = \sqrt{119 + 2} = \sqrt{121} = 11$.

Let the equation of an ellipse be $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$. Then, the radius of the circle with center $(0, \sqrt{2})$ and passing through the foci of the ellipse is

Similar Questions

If the normal at an end of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through an extremity of the minor axis,then the eccentricity $e$ of the ellipse satisfies:

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 $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.

In an ellipse,its foci and the ends of its major axis are equally spaced. If the length of its semi-minor axis is $2 \sqrt{2}$,then the length of its semi-major axis is

The centre of the ellipse $4x^2 + 9y^2 - 16x - 54y + 61 = 0$ is

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module