Let $x^2=4 k y, k>0$ be a parabola with vertex $A$. Let $B C$ be its latusrectum. An ellipse with centre on $B C$ touches the parabola at $A$, and cuts $B C$ at points $D$ and $E$ such that $B D=D E=E C(B, D, E, C$ in that order). The eccentricity of the ellipse is

Let $P$ be a variable point on the ellipse $x^2 + 3y^2 = 3$ , then the maximum perpendicular distance of $P$ from the line $x -y = 10$ is

The line $y=x+1$ meets the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then $(3 r )^{2}$ is equal to

An ellipse is drawn by taking a diameter of the circle ${\left( {x - 1} \right)^2} + {y^2} = 1$ as its semi-minor axis and a diameter of the circle ${x^2} + {\left( {y - 2} \right)^2} = 4$ is semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :

The position of the point $(1, 3)$ with respect to the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$

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