If the line $2x + y + k = 0$ is normal to the parabola $y^2 = -8x$,then the value of $k$ will be

Let $P(x, y)$ be a variable point on the parabola $y = 4x^2 + 1$. Let $Q(c, c)$ be the foot of the perpendicular drawn from $P$ to the line $y = x$. If $R(h, k)$ is the mid-point of $PQ$,then the locus of $R$ is:

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The graph of $y = f(x)$ represents a parabola whose focus has the coordinates:

For what value of $k$ does the line $2y - x + k = 0$ touch the parabola $x^2 + 4y = 0$?

If the focus of a parabola is $(3, -4)$ and its directrix is $x + y - 2 = 0$,find the vertex.

The normal at a point on the parabola $y^2 = 4x$ passes through a point $P$. Two more normals to this parabola also pass through $P$. If the centroid of the triangle formed by the feet of these three normals is $G(2,0)$,then the abscissa of $P$ is