We can simplify it to $$\sqrt {x^2+(y+4)^2}-\sqrt {x^2+(y-3)^2}=5$$ Therefore, the difference of distance of $P(x,y)$ from $(0,-4)$ and $(0,3)$ is $5$.
This probably represents of hyperbola since $PS_1-PS_2= k$ where $P$ is the moving poing and $S_1$ and $S_2$ are focii.
The answer says it’s an infinite ray. It does seem plausible, but is their an analytical way which shows that it’s a not a hyperbola
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Say we have a point $P(x,y)$ and points $A(0,-4)$ and $B(3,0)$. Now we the condition
$$\sqrt {x^2+(y+4)^2}-\sqrt {(x-3)^2+y^2}=5$$
is actually $$PA-PB =AB$$
If $B$ is not on a line $AP$ then we have by triangle inequality $$PB+BA>AB$$
So $B$ must lie on a line $AP$. So $P$ is on half line which starts at $A$ and goes throught $B$.
On
Note that the equation is,
$$\sqrt {x^2+(y+4)^2}-\sqrt {(x-3)^2+y^2}=5$$
which would be the half of the hyperbola around the focus $(3,0)$ and with 5 as the difference of distance between any point on the hyperbola to the two foci $(0,-4)$ and $(3,0)$. However, in this case, the distance between the foci is also $\sqrt{3^2+4^2} = 5$, which depresses the span of the hyperbola in the direction of the minor axis to zero, i.e. $b=\sqrt{c^2-a^2} = 0$.
Thus, it is the special case where the half hyperbola degenerates into a ray with the vertex coinciding with the focal point. The ray originates at the vertex $(3,0)$ and lies in the direction of the major axis.
Solve for the first radical and square: $$x^2+y^2+8y+16=x^2+y^2-6x+9+10\sqrt{x^2+y^2-6x+9}+25$$ Solve for the second radical and square again: $$(6x+8y-18)^2=100(x^2+y^2-6x+9)$$ Simplify: $$16x^2+9y^2-24xy-96x+72y+144=0$$ Factor: $$4(4x-3y-12)^2=0$$ So we have $$y=\frac{4x-12}3$$ But we have squared so we may have introduced extraneous roots. Substitute into the first radical: $$\sqrt{x^2+y^2+8y+16}=\sqrt{x^2+\left(\frac{4x-12}3\right)^2+8\left(\frac{4x-12}3\right)+16}=\left|\frac{5x}3\right|$$ And the second: $$\sqrt{x^2+y^2-6x+9}=\sqrt{x^2+\left(\frac{4x-12}3\right)^2-6x+9}=\left|\frac{5(x-3)}3\right|$$ So now $$\left|\frac{5x}3\right|-\left|\frac{5(x-3)}3\right|=5$$ If $x\ge3$ this is an identity. If $0<x<3$ then it reads $$\frac{5x}3+\frac{5(x-3)}3=\frac{10x}3-5=5$$ Which is only satisfied for $x=3$ which is not in the currently tested range. If $x\le0$ it reads $$-5=5$$ So it is false for $x$ in the currently tested range. The only values of $x$ that work are then $x\ge3$ so the locus is $$\left\{(x,y)\vert x\ge3\cap4x-3y=12\right\}$$