Question 1099223
{{{y=2x^2}}}
{{{x=16y^2}}}
Substituting,
{{{y=2(16y^2)^2}}}
{{{y=2(256y^4)}}}
{{{y=512y^4}}}
{{{512y^4-y=0}}}
{{{y(512y^3-1)=0}}}
Two solutions:
{{{y=0}}} where then {{{x=0}}}
and
{{{512y^3=1}}}
{{{y^3=1/512}}}
{{{y=1/8}}}} and {{{x=16(1/8)^2=1/4}}}
So the points of intersection are,
({{{0}}},{{{0}}}) and ({{{1/4}}},{{{1/8}}})
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*[illustration CK14.JPG].