If $x^2 = 8ay$ is the transformed equation of $x^2 - 4y + 6x + 15 = 0$ when the origin is shifted to the point $(\alpha, \beta)$ by translation of axes,then $2\alpha + 8\beta^2 =$

The angle between the two curves $y^2 = 4(x + 1)$ and $x^2 = 4(y + 1)$ is .............. $^\circ$.

The intercepts of the focal chord (which is a part of the latus rectum) to the parabola $y^{2}=4ax$ are $b$ and $k$. Then $k$ is equal to:

The vertex of the parabola $3x - 2y^2 - 4y + 7 = 0$ is

The equations of the sides $AB$ and $AC$ of a triangle $ABC$ are $(\lambda+1) x +\lambda y =4$ and $\lambda x +(1-\lambda) y +\lambda=0$ respectively. Its vertex $A$ is on the $y$-axis and its orthocentre is $(1,2)$. The length of the tangent from the point $C$ to the part of the parabola $y^2=6 x$ in the first quadrant is

Find the equation of the normal to the parabola $y^2 + 12x = 0$ at the upper end of its latus rectum.