Given the equation of a parabola:
[tex]5x^2-50x-4y+113=0[/tex]• You can rewrite it in Standard Form by following these steps:
1. Add the y-term to both sides of the equation:
[tex]\begin{gathered} 5x^2-50x-4y+113+(4y)=0+(4y) \\ \\ 5x^2-50x+113=4y \end{gathered}[/tex]2. Subtract the Constant Term from both sides of the equation:
[tex]\begin{gathered} 5x^2-50x+113-(113)=4y-(113) \\ \\ 5x^2-50x=4y-113 \end{gathered}[/tex]3. Divide both sides of the equation by 5 (the leading coefficient)
[tex]x^2-10x=\frac{4}{5}y-\frac{113}{5}[/tex]4. The coefficient of the x-term is:
[tex]b=-10[/tex]Then, you need to add this value to both sides:
[tex](\frac{-10}{2})^2=5^2[/tex]Therefore.
[tex]\begin{gathered} x^2-10x+5^2=\frac{4}{5}y-\frac{113}{5}+5^2 \\ \\ x^2-10x+5^2=\frac{4}{5}y+\frac{12}{5} \end{gathered}[/tex]5. Rewrite the equation as follows:
[tex]\begin{gathered} (x-5)^2=\frac{4}{5}(y+3) \\ \\ (x-5)^2=0.8(y+3) \end{gathered}[/tex]• Having the equation written in Standard Form:
[tex](x-h)^2=4p(y-k)[/tex]You can identify that:
[tex]4p=\frac{4}{5}[/tex]Solving for "p", you get:
[tex]p=\frac{4}{5\cdot4}=\frac{1}{5}=0.2[/tex]• You can identify that:
[tex]\begin{gathered} h=5 \\ k=-3 \end{gathered}[/tex]Therefore, the Vertex is:
[tex](5,-3)[/tex]• By definition, the Focus of a parabola that opens upward is given by:
[tex](h,k+p)[/tex]Then, in this case, this is:
[tex](5,-3+0.2)=(5,-2.8)[/tex]• By definition, the Directrix for a parabola that opens upward is given by:
[tex]y=k-p[/tex]Then, in this case, you get:
[tex]\begin{gathered} y=-3-0.2 \\ \\ y=-3.2 \end{gathered}[/tex]Hence, the answers are:
• Standard Form:
[tex](x-5)^2=0.8(y+3)[/tex]• Value for "p":
[tex]p=0.2[/tex]• Vertex:
[tex](5,-3)[/tex]• Focus:
[tex](5,-2.8)[/tex]• Directrix:
[tex]y=-3.2[/tex]
