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=\s

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

The eccentricity of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ is

$P$ is a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $AA'$ as the major axis. Then the maximum value of the area of $\Delta APA'$ is

In an ellipse $9x^2 + 5y^2 = 45$,the distance between the foci is

Number of points on the ellipse $\frac{x^2}{50} + \frac{y^2}{20} = 1$ from which a pair of perpendicular tangents are drawn to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

The mid-point of a chord of the ellipse $x^2+4y^2-2x+20y=0$ is $(2,-4)$. The equation of the chord is

Vedclass Test Series

Exam Paper Generator

Online Exam Module