Question

The directex of Y=1/8x^2 is?

Answer 1

`\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(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}\\\\ -------------------------------\\\\ y=\cfrac{1}{8}x^2\implies (y-0)=\cfrac{1}{8}(x-0)^2\implies 8(y-0)=(x-0)^2 \\\\\\ 4(2)(y-0)=(x-0)^2\impliedby \textit{that means, p = 2}`

so... if you notice, the vertex is at h,k and that'd be the origin, 0,0

so...since the directrix is "p" units from the vertex, so it'd be 2 units from 0,0

now, the parabola has an equation with a positive leading term's coefficient, namely the 1/8 is positive, thus, the parabola is opening upwards, and the directrix is "outside" the parabola, so is below the vertex

that puts the directrix 2 units below 0,0

y = -2