In an ellipse, the distance between its foci | vedclass

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

Question

$2 \mathrm{ae}=6 \Rightarrow \mathrm{ae}=3$

Minor axis $=2 b=8 \Rightarrow b=4$

As, $b^{2}=a^{2}\left(1-e^{2}\right) \Rightarrow 16=a^{2}-a^{2}

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

$2 \mathrm{ae}=6 \Rightarrow \mathrm{ae}=3$

Minor axis $=2 b=8 \Rightarrow b=4$

As, $b^{2}=a^{2}\left(1-e^{2}\right) \Rightarrow 16=a^{2}-a^{2} e^{2}$

$\Rightarrow 16=a^{2}-9$

$\Rightarrow \mathrm{a}^{2}=25 \Rightarrow \mathrm{a}=5$

ae $=3 \Rightarrow \mathrm{e}=\frac{3}{5}$

In an ellipse, the distance between its foci is $6$ and minor axis is $8.$ Then its eccentricity is :

Similar Questions

The number of values of $c$ such that the straight line $y = 4x + c$ touches the curve $\frac{{{x^2}}}{4} + {y^2} = 1$ is

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1) $ and has eccentricity $\sqrt {\frac{2}{5}} $ is

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

If the normal at any point $P$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ meets the co-ordinate axes in $G$ and $g$ respectively, then $PG:Pg = $