Answer :
Answer:
Step-by-step explanation:
[tex]2y+18x=2x\left(x+9\right)\\Parabola\:standard\:equation\\4p\left(yk\right)=\left(xh\right)^2\:\mathrm{\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:parabola\:with\:vertex\:at}\:\left(h,\:k\right), \\\mathrm{and\:a\:focal\:length\:}\:|p|\\\mathrm{Rewrite}\:2y+18x=2x\left(x+9\right)\:\mathrm{in\:the\:standard\:form}\\2y+18x=2x\left(x+9\right)\\\mathrm{Rewrite\:as}\\2y=2x\left(x+9\right)-18x\\\mathrm{Divide\:by}\:2\\y=x^2\\\mathrm{Rewrite\:in\:standard\:form}[/tex][tex]4\cdot \frac{1}{4}\left(y-0\right)=\left(x-0\right)^2\\Therefore\:parabola\:properties\:are:\\\left(h,\:k\right)=\left(0,\:0\right),\:p=\frac{1}{4}[/tex]