Mathematics, 28.07.2019 08:30, marcusajns
(03.09 mc) derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. put the equation in standard form
Answers: 2
![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2 \\\\\\ -4(y-5)=(x+2)^2\implies y-5=-\cfrac{1}{4}(x+2)^2 \\\\\\ y=-\cfrac{1}{4}(x+2)^2+5](/tpl/images/0142/1921/74bf3.png)
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(03.09 mc) derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. put...
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