The locus of the orthocentre of the triangle formed by the focal chord of the parabola $y^2 = 4ax$ and the normals drawn at its extremities is

If the normal chord drawn at $(2 a, 2 a \sqrt{2})$ on the parabola $y^2=4 a x$ subtends an angle $\theta$ at its vertex,then $\theta=$ (in $^{\circ}$)

$A$ point $P$ moves such that the sum of the angles which the three normals drawn from $P$ to the standard parabola $y^2 = 4ax$ make with the axis of the parabola is constant. Then the locus of $P$ is:

If the normal at the point $t_1$ (i.e.,at $(at_1^2, 2at_1)$) on the parabola $y^2 = 4ax$ meets the parabola again at the point $t_2$,then $t_1t_2$ is equal to:

If $b$ and $c$ are the lengths of the segments of any focal chord of a parabola $y^2 = 4ax$,then the length of the semi-latus rectum is:

$A$ line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan^{-1}\left(\frac{3}{4}\right)$ with the $X$-axis. It intersects the parabola $y^{2}=4(x-3)$ at points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively. Then $|x_{1}-x_{2}|$ is equal to