Answer
$(0,-3-2\sqrt{5}) $ and $ (0,-3+2\sqrt{5})$
Work Step by Step
A point on the $y$ axis has coordinates $P_{1}(0,y).$
Given $P_{2}(4,-3)$, we find $x$ from
$ \begin{aligned}&\\
d(P_{1},P_{2})&=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\
6&=\sqrt{(4-0)^{2}+(-3-y)^{2}}\\
6^{2}&=(4-0)^{2}+(-3-y)^{2}\\
36&=16+9+6y+y^{2}\\
0&=y^{2}+6y-11
\end{aligned}$
Using the quadratic formula,
$y=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=$
$=\displaystyle \frac{-6\pm\sqrt{36+44}}{2}=$
$=\displaystyle \frac{-6\pm\sqrt{80}}{2}$
$=\displaystyle \frac{-6\pm 4\sqrt{5}}{2}$
$=-3\pm 2\sqrt{5}$
The points have coordinates $(0,-3-2\sqrt{5}) $ and $ (0,-3+2\sqrt{5})$
