If the vertices of a triangle are $(am_1^2, 2am_1), (am_2^2, 2am_2),$ and $(am_3^2, 2am_3),$ then the area of the triangle 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

Let $P$ be a point on the parabola $y^2 = 4ax$,where $a > 0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $PFQ$,where $F$ is the focus of the parabola,is $120$. If the slope $m$ of the normal and $a$ are both positive integers,then the pair $(a, m)$ is

If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(2, 3)$ to the parabola $y^2 = 4x$,then what is the value of $\frac{1}{m_1} + \frac{1}{m_2}$?

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