What is the equation of the ellipse with foci $( \pm 2,\;0)$ and eccentricity $ = \frac{1}{2}$
$3{x^2} + 4{y^2} = 48$
$4{x^2} + 3{y^2} = 48$
$3{x^2} + 4{y^2} = 0$
$4{x^2} + 3{y^2} = 0$
Let $A = \left\{ {\left( {x,y} \right):\,y = mx + 1} \right\}$
$B = \left\{ {\left( {x,y} \right):\,\,{x^2} + 4{y^2} = 1} \right\}$
$C = \left\{ {\left( {\alpha ,\beta } \right):\,\left( {\alpha ,\beta } \right) \in A\,\,and\,\,\left( {\alpha ,\beta } \right) \in B\,\,and\,\alpha \, > 0} \right\}$ .
If set $C$ is singleton set then sum of all possible values of $m$ is
The point $(4, -3)$ with respect to the ellipse $4{x^2} + 5{y^2} = 1$
Eccentric angle of a point on the ellipse ${x^2} + 3{y^2} = 6$ at a distance $2$ units from the centre of the ellipse is
The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length $4$ and distance between foci $4 \sqrt 2$ is-
In an ellipse, the distance between its foci is $6$ and minor axis is $8.$ Then its eccentricity is :