The equation of the tangent to the parabola $y^2=16x$,which is perpendicular to the line $3x-4y+5=0$,is given by

Let $ABCD$ be a trapezium whose vertices lie on the parabola $y^2=4x$. Let the sides $AD$ and $BC$ of the trapezium be parallel to the $y$-axis. If the diagonal $AC$ is of length $\frac{25}{4}$ and it passes through the point $(1,0)$,then the area of $ABCD$ is:

$A$ normal chord of the parabola $y^2 = 4x$ subtending a right angle at the vertex makes an acute angle $\theta$ with the $x$-axis. Then $\theta$ is equal to:

From the focus of the parabola $y^2 = 12x$, a ray of light is directed in a direction making an angle $\tan^{-1} \frac{3}{4}$ with the $x$-axis. Then the equation of the line along which the reflected ray leaves the parabola is

If the graph of $y = ax^2 - bx + c$ is as shown below,then the signs of $a$,$b$,and $c$ are:

If one end of a focal chord $AB$ of the parabola $y^{2}=8x$ is at $A\left(\frac{1}{2},-2\right)$,then the equation of the tangent to it at $B$ is: