The writing is almost automatic:
$$\sqrt{(x-4)^2+(y+3)^2}=\frac{1}{2}\sqrt{(x+1)^2+(y+1)^2}.$$
You will find it worthwhile to simplify. Square both sides. After some manipulation, you will find that the locus is a circle, of which you can find the centre and radius.
Remark: If what I called almost automatic is not quite as automatic as I think, please leave a message and I can amplify.
I think you overworked here. The condition is
$$|AP|^2=9|BP|^2\iff (x+3a)^2+y^2=9\left[(x-a)^2+y^2\right]\iff$$
$$\iff 8x^2-24ax+8y^2=0\iff \left(x-\frac{3a}2\right)^2+y^2=\frac{9a^2}4$$
and we get that the points$\;P=(x,y)\;$ are the locus of circle with center $\;\left(\frac{3a}2\,,\,0\right)\;$ and radius $\;\frac{3a}2\;$ , and since radius = absolute value of $\;x$ - coordinate of circle, the circle is tangent to the $\;y$ - axis
Let $\;Q=(x,y)\;$ , then its distance from the $\;y$ - axis is $\;|x|\;$ , and the length of any of its two tangets to the above circle is
$$\sqrt{\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4}$$
and thus we get the equation:
$$x^2=\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4\iff$$
$$\iff -3ax+y^2=0\;\;(*)$$
Now, let us check what is the locus of all point whose distance to $\;4x+3a=0\iff x=-\frac{3a}4\;$ equals its distance to $\;\left(\frac{3a}4\,,\,0\right)\;$:
$$\left(x-\frac{3a}4\right)^2+y^2=\frac{(4x+3a)^2}{16}\iff x^2-\frac{3a}2x+\frac{9a^2}{16}+y^2=x^2+\frac{3a}2x+\frac{9a^2}{16}\iff$$
$$\iff -3ax+y^2=0\;\;(*)$$
and thus both conditions are identical
Best Answer
a) The distance of the point $P(x, y)$ from the point $A(3, 1)$ is such that $$\sqrt{(x-3)^2+(y-1)^2}=3\times \frac{|x+1|}{\sqrt{1^2+(0)^2}}$$ $$(x-3)^2+(y-1)^2=9(x+1)^2$$ $$x^2-6x+9+y^2-2y+1=9(x^2+2x+1)$$ $$8x^2-y^2+24x+2y-1=0$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the locus of point P:}\ 8x^2-y^2+24x+2y-1=0}}$$
Above is the locus of the point $P(x, y)$
b) setting $y=-1$ in the above equation, we get $$8x^2-(-1)^2+24x+2(-1)-1=0$$ $$2x^2+6x-1=0$$ Checking the nature of roots of above quadratic equation by using discriminant $\Delta$, as follows $$\Delta=B^2-AC=(6)^2-4(2)(-1)=44>0$$ Above positive value of the discriminant shows that there are two distinct real roots i.e. the locus of the point P intersects the straight line: $y=-1$ at two different points .