Find the equation of the chord of the parabola $y^2 = 6x$ passing through the vertex and the negative end of the latus rectum.

Let $L_{1}$ be a tangent to the parabola $y^{2}=4(x+1)$ and $L_{2}$ be a tangent to the parabola $y^{2}=8(x+2)$ such that $L_{1}$ and $L_{2}$ intersect at right angles. Then $L_{1}$ and $L_{2}$ meet on the straight line

If the point on the curve $y^{2}=6x$,nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$,then $2(\alpha+\beta)$ is equal to $.....$

In which quadrant does the vertex of the parabola $y^2 + 2y + x = 0$ lie?

If $(-1,-1)$ is the focus and $x+y+4=0$ is the directrix of a parabola,then its vertex is

If the normal at one end of the latus rectum of the parabola $y^2=16x$ meets the $X$-axis at the point $P$, then the length of the chord passing through $P$ and perpendicular to the normal is (in $\sqrt{2}$)