$A$ 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

The tangents to the parabola $y^2 = 4ax$ from an external point $P$ make angles $\theta_1$ and $\theta_2$ with the axis of the parabola,such that $\tan \theta_1 + \tan \theta_2 = b$,where $b$ is a constant. Then $P$ lies on

Three points $O(0,0)$,$P(a, a^2)$,and $Q(-b, b^2)$ with $a > 0$ and $b > 0$ lie on the parabola $y = x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola,and $S_2$ be the area of the triangle $OPQ$. If the minimum value of $\frac{S_1}{S_2}$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to:

The line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = x + 2$ at the points $P$ and $Q$. If the coordinates of the point $X$ are $(\sqrt{3}, 0)$,then the value of $XP \cdot XQ$ is

The coordinates of the extremities of the latus rectum of the parabola $5y^2 = 4x$ are

If $x = t^{2} + 2$ and $y = 2t$ represent the parametric equations of a parabola,find its Cartesian equation.