Question

I need to know 2 idk how to do this, help please

Answer 1

so, the squared variable, is the "y", that means, the parabola is opening horizontally, over the x-axis
now, the leading term's coefficient, is positive, that means it opens to the right-hand-side, like the one in the picture below


`\bf \begin{array}{llll} \boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\ (x-{{ h}})^2=4{{ p}}(y-{{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -----------------------------\\\\ \begin{array}{llccll} x=&\cfrac{1}{2}(y&-2)^2&+3\\ &\uparrow &\uparrow &\uparrow \\ &4p&k&h \end{array} \qquad now\qquad 4p=\cfrac{1}{2}\implies p=\cfrac{1}{8}`

so, we know the distance "p" is 1/8, the h,k are 3 and 2, respectively

so the vertex is at 3,2 and the focus is 1/8 from there to the right
and the directrix is 1/8 from there, to the left

focus 3+1/8 = 25/8
directrix 3 - 1/8 = 23/8

since the vertex is 3,2, and is running horizontally, the axis of symmetry will be y = 2

the latus rectum, or "focal width", will just be 4p, or 4*1/8