The locus of a point which moves so that it is always equidistant from the points $A(a, 0)$ and $B(-a, 0)$ is

The locus of a point $P$ which moves such that the sum of its distances from two perpendicular lines is equal to $1$ is a

Let $a \neq 0, b \neq 0, c$ be three real numbers and $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}, \forall p, q \in \mathbb{R}$. If $L\left(\frac{2}{3}, \frac{1}{3}\right) + L\left(\frac{1}{3}, \frac{2}{3}\right) + L(2, 2) = 0$,then the line $ax + by + c = 0$ always passes through the fixed point:

If $A=(-1, 2)$ and $B=(1, -2)$ are two points and $P$ is a variable point such that the area of $\triangle PAB$ is always $1$,then the equation of the locus of $P$ is

If for a variable line $\frac{x}{a} + \frac{y}{b} = 1$,the condition $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$ ($c$ is a constant) is satisfied,then the locus of the foot of the perpendicular drawn from the origin to the line is:

$A$ rod of length $8$ units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$,respectively. If the locus of the point $P$,that divides the rod $AB$ internally in the ratio $2:1$ is $9(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76=0$,then $\alpha-\beta-\gamma$ is equal to :