Question 843717
Intersection of an ellipse and a parabola.
1.{{{x^2=y+10}}}
2.{{{y^2+5(x-1)^2=81}}}
From eq. 1,
{{{y=x^2-10}}}
{{{y^2=x^4-20x^2+100}}}
Substituting into eq. 2,
{{{x^4-20x^2+100+5(x^2-2x+1)=81}}}
{{{x^4-20x^2+100+5x^2-10x+5=81}}}
{{{(x-1)(x-4)(x+2)(x+3)=0}}}
.
.
Four solutions:
{{{x-1=0}}}
{{{x=1}}}
Then,
{{{y=1^2-10=-9}}}
(1,-9)
and
{{{x-4=0}}}
{{{x=4}}}
Then,
{{{y=4^2-10=6}}}
(4,6)
and
{{{x+2=0}}}
{{{x=-2}}}
Then,
{{{y=(-2)^2-10=-6}}}
(-2,-6)
and
{{{x+3=0}}}
{{{x=-3}}}
Then
{{{y=(-3)^2-10=-1}}}
(-3,-1)
{{{drawing(300,300,-9,9,-10,8,circle(4,6,0.4),circle(-2,-6,0.4),circle(1,-9,0.4),circle(-3,-1,0.4),graph(300,300,-9,9,-10,8,x^2-10,sqrt(81-5(x-1)^2),-sqrt(81-5(x-1)^2)))}}}