What is the vertex of the parabola $x^2 + 4x + 2y - 7 = 0$?

If the line $x + \alpha y + \beta = 0$ touches the curve $4x^3 + 4y^3 = xy(xy + 16)$ at points $(x_1, y_1)$ and $(x_2, y_2)$,where $x_1 \neq x_2$,then the value of $\alpha + \beta$ is (given $\alpha, \beta \in R$).

$A$ beam is supported at its ends by supports which are $12 \, m$ apart. Since the load is concentrated at its centre,there is a deflection of $3 \, cm$ at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection $1 \, cm$?

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 $.....$

The point on the curve $x^{2}=2 y$ which is nearest to the point $(0,5)$ is

$A$ normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0, -\alpha)$ to the parabola $x^2 = -4ay$,where $a > 0$. Let $L$ be the line passing through $(0, -\alpha)$ and parallel to the directrix of the parabola. Suppose that $L$ intersects the parabola at two points $A$ and $B$. Let $r$ denote the length of the latus rectum and $s$ denote the square of the length of the line segment $AB$. If $r : s = 1 : 16$,then the value of $24a$ is. . . .