If a normal chord at a point $t$ on the parabola $y^2=4ax$ subtends a right angle at the vertex,then $t^2$ equals to

In the $XY$-plane,three distinct lines $l_1, l_2, l_3$ concur at a point $(\lambda, 0)$. Further,the lines $l_1, l_2, l_3$ are normals to the parabola $y^2=6x$ at the points $A=(x_1, y_1)$,$B=(x_2, y_2)$,and $C=(x_3, y_3)$ respectively. Then,we have:

The area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the extremities of its latus rectum is

If $ax + by = 1$ is a normal to the parabola $y^2 = 4px$, then the condition is:

Let the curve $C$ be the mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$. If $A$ and $B$ are the points of intersection of $C$ with the line $y=-5$,then the distance between $A$ and $B$ is

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