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Equation of a parabola with a vertex of (1,3) and a focus of (1,5)

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so hmm notice the picture below

the focus is above the vertex, meaning the parabola is opening upwards

now... notice how far up is the focus point, from the vertex, that'd be the distance "p"

the parabola is going vertical, that means the squared variable is the "x"

thus
\bf (x-{{ h}})^2=4{{ p}}(y-{{ k}}) \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}

you know what h,k are, those are given... and now you know what "p" distance is, so .. .just plug those in, and simplify and solve for "y"
Equation of a parabola with a vertex of (1,3) and a focus of (1,5)-example-1
User Patrick Reiner
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