Question 371715
This is the intersection of a line with an ellipse.
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1.{{{(4x^2)+(9y^2)=16}}}
2.{{{2x+3y=4}}}
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From eq. 2,
{{{2x=4-3y}}}
{{{4x^2=(4-3y)^2}}}
{{{4x^2=16-24y+9y^2}}}
Substitute into eq. 1,
{{{(16-24y+9y^2)+9y^2=16}}}
{{{18y^2-24y=0}}}
{{{6y(3y-4)=0}}}
Two solutions:
{{{6y=0}}}
{{{highlight(y=0)}}}
Then from eq. 2,
{{{2x+0=4}}}
{{{highlight(x=2)}}}
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{{{3y-4=0}}}
{{{3y=4}}}
{{{highlight_green(y=4/3)}}}
Then,
{{{2x+3(4/3)=4}}}
{{{2x+4=4}}}
{{{2x=0}}}
{{{highlight_green(x=0)}}}
The two intersection points are ({{{2}}},{{{0}}}) and ({{{0}}},{{{4/3}}}).
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{{{drawing(300,300,-4,4,-4,4,grid(1),circle(2,0,0.23),circle(0,4/3,0.23),graph(300,300,-4,4,-4,4,0,sqrt((16-4x^2)/9)),graph(300,300,-4,4,-4,4,0,-sqrt((16-4x^2)/9),(4-2x)/3)))}}}