JEE MAIN Mathematics Conic Sections Parabola MCQ Previous Year Questions

Let the parabola $y=x^2+px+q$ passing through the point $(1,-1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2+q^2$ is:

Let the focal chord $PQ$ of the parabola $y^2=4x$ make an angle of $60^{\circ }$ with the positive $x$-axis, where $P$ lies in the first quadrant. If the circle, whose one diameter is $PS$, $S$ being the focus of the parabola, touches the $y$-axis at the point $(0,\alpha )$, then $5{\alpha }^2$ is equal to: