The equation of an ellipse whose eccentricity is $1/2$ and the vertices are $(4, 0)$ and $(10, 0)$ is
$3{x^2} + 4{y^2} - 42x + 120 = 0$
$3{x^2} + 4{y^2} + 42x + 120 = 0$
$3{x^2} + 4{y^2} + 42x - 120 = 0$
$3{x^2} + 4{y^2} - 42x - 120 = 0$
In an ellipse, the distance between its foci is $6$ and minor axis is $8.$ Then its eccentricity is :
If $PQ$ is a double ordinate of hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ such that $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies
Consider an elIipse, whose centre is at the origin and its major axis is along the $x-$ axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is $6$, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is
The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is
If the variable line $y = kx + 2h$ is tangent to an ellipse $2x^2 + 3y^2 = 6$ , then locus of $P(h, k)$ is a conic $C$ whose eccentricity equals