From a point $(d, 0)$, three normals are drawn to the parabola $y^{2} = x$. Then:

Let $O$ be the origin and $A$ be a point on the curve $y^2=4x$. Then the locus of the midpoint of $OA$ is:

If one of the vertices of an equilateral triangle inscribed in the parabola $y^2=12x$ coincides with the vertex of the parabola, then the area (in sq. units) of that triangle is (in $\sqrt{3}$)

The chord $PQ$ of the parabola $y^2 = x$,where one end $P$ of the chord is at point $(4, -2)$,is perpendicular to the axis of the parabola. Then the slope of the normal at $Q$ is

The length of the latus rectum of the conic $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$ is

For the parabola $y^2+6y-2x+5=0$:
$(I)$ The vertex is $(-2,-3)$
$(II)$ The directrix is $y+3=0$
Which of the following is correct?