The equation of the diameter of the parabola $y^2 = x$ corresponding to the chord $x - y + 1 = 0$ is

Find the equation of the parabola that satisfies the following conditions: Focus $(6, 0)$,directrix $x = -6$.

Let one root of the quadratic equation in $x$: $(k^2 - 15k + 27)x^2 + 9(k-1)x + 18 = 0$ be twice the other. Then the length of the latus rectum of the parabola $y^2 = 6kx$ is equal to:

If the vertex of the conic $y^{2}-4y=4x-4a$ always lies between the straight lines $x+y=3$ and $2x+2y-1=0$, then:

If the parabola $y = \alpha x^{2} - 6x + \beta$ passes through the point $(0, 2)$ and has its tangent at $x = \frac{3}{2}$ parallel to the $X$-axis,then:

Let the curve $C$ be the mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$. If $A$ and $B$ are the points of intersection of $C$ with the line $y=-5$,then the distance between $A$ and $B$ is