Chapter 9 Quadratic Relations and Conic Sections - Chapter Review - Page 672: 35

$(1,2),(16,8)$

$y^2=4x\\0.5y^2=2x$ Thus plugging this into the second equation we get: $0.5y^2-5y+8=0\\y^2-10y+16=0\\(y-2)(y-8)=0$ Thus $y=2$ or $y=8$. $x=0.25y^2$. Thus if $y=2$, then $x=1$, and if $y=8$, then $x=16$. Thus the solutions are: $(1,2),(16,8)$