I want to deduce the formula of a parabola with $(h,k)$ vertex from a graph. So far I only could deduce the formula of one with $(0,0)$ vertex, by doing this:
$\color{red}{p1}:\, D=D'$
$\color{red}{p2}:\, \sqrt{x^2+(y-c)^2}=y+c$
$\color{red}{p3}:\, x^2+(y-c)^2=(y+c)^2$
$\color{red}{p4}:\, x^2+y^2-2yc+c^2=y^2+2yc+c^2$
$\color{red}{\therefore} \, x^2=4cy$
I already built a graph for the one with $(h,k)$ vertex:
But, so far, I have only come up with this deduction, and I don't even know if it's right
$\color{red}{p1}:\, D=D'$
$\color{red}{p2}:\, \sqrt{(x-h)^2+(y-2c-k)^2}=y-k+c$
I already know that the formula is $\mathbf{(x-h)^2=4c(y-k)}$, but I want to know how one gets there. So how can I finish the deduction that I have so far, if what it has is correct?
Thanks in advance.
If you make the translation of axes represented by the equations $x'=x-h$ and $y'=y-k$, then a parabola with vertex $(x,y)=(h,k)$ becomes a parabola with vertex $(x',y')=(0,0)$ in the $x',y'$-coordinate system, whose formula, $x'^2=4cy'$, you have already deduced. Consequently, $$(x-h)^2=4c(y-k)$$ is the equation of the parabola in the $x,y$-coordinate system.
Added: In the second picture the difference in the $y$ coordinates of $P$ and $F$ is $y-c-k$ instead of $y-2c-k$. So your last equation should be \begin{equation*} \sqrt{(x-h)^{2}+(y-c-k)^{2}}=y-k+c. \end{equation*}
Then \begin{eqnarray*} (x-h)^{2}+(y-k-c)^{2} &=&\left( y-k+c\right) ^{2} \\ (x-h)^{2}+(y-k)^{2}-2c(y-k)+c^{2} &=&(y-k)^{2}+2c(y-k)+c^{2} \\ (x-h)^{2}-2c(y-k) &=&2c(y-k) \\ (x-h)^{2} &=&4c(y-k), \end{eqnarray*} as above.