What is the equation of the locus of a point which moves such that $4$ times its distance from the $x$-axis is the square of its distance from the origin?

The side $AB$ of $\triangle ABC$ is fixed and is of length $2a$ units. The vertex $C$ moves in the plane such that the vertical angle $\angle ACB$ is always constant and is equal to $\alpha$. Let the $x$-axis be along $AB$ and the origin be at $A$. Then the locus of the vertex $C$ is:

The locus of the point of intersection of perpendicular tangents to the circle $x^{2}+y^{2}=16$ is

Let $P(\alpha, \beta)$ be a variable point which moves in the $x-y$ plane such that $\frac{PA}{PB} = 2$,where $A(1, 0)$ and $B(0, -1)$. If $M$ and $m$ denote respectively the maximum and minimum value of $\alpha + \beta$,then the value of $[\frac{M}{m}]$ is- (where $[.]$ denotes the greatest integer function)

For any value of $\theta$,if the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ intersect at $P(\theta)$,then the locus of $P(\theta)$ is a

The equation $r \cos \theta = 2 a \sin^2 \theta$ represents the curve