Two mutually perpendicular tangents of the parabola $y^2 = 4ax$ meet the axis in $P_1$ and $P_2$. If $S$ is the focus of the parabola,then $\frac{1}{SP_1} + \frac{1}{SP_2}$ is equal to

Let $x^2=4ky, k>0$ be a parabola with vertex $O(0,0)$. Let $BC$ be its latus rectum. An ellipse with center $P$ on $BC$ touches the parabola at $O$,and cuts $BC$ at points $D$ and $E$ such that $BD=DE=EC$ ($B, D, E, C$ in that order). The eccentricity of the ellipse is

The locus of the foot of the perpendicular from the centre of the hyperbola $xy = c^2$ on a variable tangent is :

If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is $1$,then $b^2=$

$TP$ and $TQ$ are tangents to the parabola $y^2 = 4ax$ at $P$ and $Q$. If the chord $PQ$ passes through the fixed point $(-a, b)$,then the locus of $T$ is:

$A$ common tangent $T$ to the curves $C_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$ does not pass through the fourth quadrant. If $T$ touches $C_{1}$ at $(x_{1}, y_{1})$ and $C_{2}$ at $(x_{2}, y_{2})$,then $|2x_{1} + x_{2}|$ is equal to $......$