A parabola with focus $(3,4)$ touches $y=x$ and $y=0$. Find its directrix, equation of parabola and vertex. The following problem is taken from the practice set of JEE exam.

A parabola with focus $(3,4)$ touches $y=x$ and $y=0$. Find its directrix, equation of parabola and vertex.

I could not picture this parabola. I thought the statement was wrong. But in the hint they have written this:

Directrix is the line joining the image of $(3,4)$ in $y=x$ and $y=0$.

Image of $(3,4)$ in $y=x$ is $(4,3)$ and in $y=0$ is $(3,-4)$. Equation of line through these two points is $7x-y-25=0$
My doubt is: why is directrix obtained by joining the image of focus about two tangents?
Do we need to use here the property that the tangents at the extremities of a focal chord of a parabola intersect at right angle on the directrix? But not sure how to use this here.
 A: Let $F = (3, 4)$ and let $P$ be the point of tangency between the parabola and a tangent line.  The reflection of $F$ about the tangent line produces $F'$ that is has
the property that $|FP|= |F'P|$ and also we have that $PF'$ is orthogonal to the directrix (because it is parallel to the axis of the parabola), therefore $F'$ lies on the directrix.  Two such lines determine two images of the focus that both lie on the same line.  Therefore the direction is just the line connecting the two focus images.
The reflection image of $(3, 4)$ across $y = x$ is $(4, 3)$ and across $y = 0$ is $(3, -4)$, therefore the equation of the directrix is
$  y = 3 + 7 (x - 4) = 7 x - 25 $
The equation of the parabola is
$ (x - 3)^2 + (y - 4)^2 = ( 7 x - y - 25 )^2 / (7^2 + 1^2) $
Distance of $(3, 4)$ from the directrix is $(21 - 4 - 25)/\sqrt{50} = \dfrac{8}{5\sqrt{2} } $
The axis is along the vector $(7, -1)$, so the vertex is given by
$ (3, 4) + \dfrac{4}{50} (7, -1) =  (3 + \dfrac{14}{25}, 4 - \dfrac{2}{25} ) = (\dfrac{89}{25}, \dfrac{98}{25} )$
