Question 884679
Ellipse,   {{{4x^2+9y^2=144}}}
.                                                                                 Line,   {{{2x+3y=12}}}
.


You can identify the type of conic section just by looking at those equations.  The ellipse equation is almost but not yet in standard form and the line is in standard form.  Identification is uncomplicated.


Solving the system of the two equations is a little more work.
Solve the linear equation for either variable and substitute into the ellipse equation and solve for the single variable.  You could use the linear equation again to determine the other variable.


{{{3y=-2x+12}}}
{{{y=-2x/3+4}}}
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{{{4x^2+9(-2x/3+4)^2=144}}}
{{{4x^2+9((4/9)x^2-16x/3+16)=144}}}
{{{4x^2+4x^2-3*16x+9*16=144}}}
{{{8x^2-48x+151=144}}}
{{{8x^2-48x+7=0}}}
Discriminant:  {{{(-48)^2-4*8*7=2080}}}
Factoring Discriminant:  {{{208*10=104*2*10=52*2*2*10=4*13*2*2*2*10=2^5*13*2*5=2^6*5*13}}}.
Continuing with general solution of quadratic equation:
{{{x=(48+- sqrt(2^6*5*13))/(2*8)}}}
{{{x=(48+- 8sqrt(65))/(2*8)}}}
{{{highlight(x=(6+- sqrt(65))/2)}}}, for the two x coordinates.


FINDING y COORDINATES
Found from earlier, {{{y=-2x/3+4}}}.
{{{y=-(2/3)x+4}}}
{{{y=-(2/3)((6+- sqrt(65))/2)+4}}}
{{{y=-4+- (1/3)sqrt(65)+4}}}, notice the resulting additive inverses 4 and -4.
{{{highlight(y=-(1/3)sqrt(65))}}}  OR  {{{highlight((1/3)sqrt(65))}}}.



Careful which coordinate goes with which other coordiante.
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x,y:    {{{6-sqrt(65)/2}}}, {{{sqrt(65)/3}}}
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x,y:  {{{6+sqrt(65)/2}}}, {{{-sqrt(65)/3}}}