The focus is $$F(4,-1)$$ and the tangent at the vertex is $$3x+4y-11=0$$
The perpendicular distance from the focus to this line is $$\frac{|3(4)+4(-1)-11|}{\sqrt{3^2+4^2}}$$
$$=\frac{|12-4-11|}{5}$$
$$=\frac35$$
So,
$$a=\frac35$$
The directrix is parallel to $$3x+4y-11=0$$
Let its equation be $$3x+4y+k=0$$
The line through the focus parallel to the tangent is
$$3x+4y-8=0$$
Since the tangent at the vertex lies midway between this line and the directrix,
$$-11=\frac{-8+k}{2}$$
$$-22=-8+k$$
$$k=-14$$
Therefore, the directrix is
$$3x+4y-14=0$$
Now, any point
$$P(x,y)$$ on the parabola satisfies
$$PF=PM$$
Hence,
$$\sqrt{(x-4)^2+(y+1)^2}=\frac{|3x+4y-14|}{5}$$
Squaring,
$$25\left[(x-4)^2+(y+1)^2\right]=(3x+4y-14)^2$$
Expanding,
$$25(x^2-8x+16+y^2+2y+1)=9x^2+16y^2+196+24xy-84x-112y$$
$$25x^2+25y^2-200x+50y+425=9x^2+16y^2+24xy-84x-112y+196$$
Bringing all terms to one side,
$$16x^2-24xy+9y^2-116x+162y+229=0$$
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