Focal Property

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,

$$16x^2-24xy+9y^2-116x+162y+229=0$$