The vertex of the parabola $(y - 2)^2 = 16(x - 1)$ is

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 straight line $y + x = 1$ touches the parabola:

The tangent to the parabola $y^2 = 4ax$ at the point $(a, 2a)$ makes an angle with the $x$-axis equal to

Let $P$ be a point on the parabola $x^2 = 4y$. If the distance of $P$ from the centre of the circle $x^2 + y^2 + 6x + 8 = 0$ is minimum,then the equation of the tangent to the parabola at $P$ is:

Tangents are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$. The length of the intercept made by these tangents on the line $x = 2$ is: