What is the equation of the tangent to the parabola $y^2 = 4x$ at the point $(1, 2)$?

Find the point where the normal at the point $(4, 4)$ intersects the parabola $y^2 = 4x$ again.

The vertex of the parabola $y^2 + 2y + x = 0$ lies in which quadrant?

$A$ 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:

Two tangents are drawn from a point $(-2, -1)$ to the curve $y^2 = 4x$. If $\alpha$ is the angle between them,then $|\tan \alpha|$ is equal to:

If a normal chord of a parabola $y^2 = 4ax$ subtends a right angle at the origin,then the slope of that normal chord is