Question

What is the focus of the parabola given by -1/4(y+2)^2=(x-1)

Answer 1

`\bf \textit{parabola vertex form with focus point distance}\\\\ \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}\\\\ -------------------------------\\\\`


`\bf -\cfrac{1}{4}(y+2)^2=(x-1)\implies (y+2)^2={-4}(x-1) \\\\\\\ [y-(-2)]^2=-4(x-1)\quad \begin{cases} k=-2\\ h=1\\ 4p=-4 \end{cases}\implies 4p=-4\implies \boxed{p=-1}`

so, is a horizontal parabola, the "p" distance is 1, however, we ended up with a negative value, that means, the parabola is opening to the left-hand-side, with a vertex at (1, -2), and its focus at 0, -2, like you see in the picture below, one unit to the left of the vertex.