The vertex of the parabola is at $(1, 2)$ and its axis is parallel to the $y$-axis. If the parabola passes through $(0, 6)$,then its latus rectum is:

The equation of the given curve is $x^2-4x+4y-8=0$. Match the following:

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

The correct matching is:

Let $PQ$ be a chord of the parabola $y^2=12x$ and the midpoint of $PQ$ be at $(4,1)$. Then,which of the following points lies on the line passing through the points $P$ and $Q$?

If $x^2 = 8ay$ is the transformed equation of $x^2 - 4y + 6x + 15 = 0$ when the origin is shifted to the point $(\alpha, \beta)$ by translation of axes,then $2\alpha + 8\beta^2 =$

The parametric equations of the curve $y^2 = 8x$ are

If $y=mx+1$ is a tangent to the parabola $y^2=4x$,then $m=$