To find the intersection point(s) between a parabola and a conic, we solve one or more system(s) of second-degree equations.
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Use the appropriate method (comparison, substitution, or elimination) to obtain an equation with one variable.
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Manipulate the equation so that it equals |0.|
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Solve the equation to find the value(s) of the isolated variable.
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Replace the value(s) obtained in one of the original equations to obtain the value(s) of the other variable.
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Write the coordinates of the point(s) of intersection.
Unlike the intersection between a line and a conic, there are five possible cases regarding the number of solutions:
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the parabola and the conic do not intersect;
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the parabola and the conic intersect only at one place, called the point of tangency; or
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the parabola and the conic intersect in two, three, or four distinct places.
In the following interactive video, select a conic section and drag the cursor to study possible cases.
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Use the appropriate method to obtain a one-variable equation
Use the substitution method since |y^2| in the parabola’s equation is already isolated.||x2+y2=36x2+−16x=36|| -
Manipulate the equation so that it equals |0|
||x2−16x=361x2−16x−36=0|| -
Solve the equation
Use the quadratic formula to obtain:||x=−b±√b2−4ac2a=−(−16)±√(−16)2−4(1)(−36)2(1)=16±√4002x∈{−2,18}||The equation shows that the parabola is horizontal and open to the left. Thus, reject |x=18,| since the function does not exist at this value.||x=-2|| -
Substitute the values obtained in one of the starting equations
||y2=−16x=−16(−2)=32y=±√32yA≈5.66yB≈−5.66|| -
Write the coordinates of the points of intersection
The coordinates of the two points of intersection between the parabola |y^2=-16x| and the circle |x^2+y^2=36| are |A(-2,5{.}66)| and |B(-2,-5{.}66).|

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Use the appropriate method to obtain a one-variable equation
Use the substitution method, since |x^2| in the parabola’s equation is already isolated.||x225+y29=112(y+3)25+y29=1|| -
Manipulate the equation so that it equals |0|
||12y+3625+y29=19(12y+36)225+25y2225=1108y+324+25y2225=125y2+108y+324=22525y2+108y+99=0|| -
Solve the equation
Use the quadratic formula to obtain: ||y=−b±√b2−4ac2a=−(108)±√(108)2−4(25)(99)2(25)=−108±√1 76450y∈{−3,−1.32}|| -
Substitute the values obtained in one of the starting equations
For |y=-3|: ||x2=12(y+3)=12(−3+3)=0xA=0||For |y=-1{.}32|:||x2=12(y+3)=12(−1.32+3)=20.16x=±√20.16xB≈−4.49xC≈4.49|| -
Write the coordinates of the points of intersection
The coordinates of the three points of intersection between the parabola |x^2=12(y+3)| and the ellipse |\dfrac{x^2}{25}+\dfrac{y^2}{9}=1| are |A(0,-3),| |B(-4{.}49,-1{.}32),| and |C(4{.}49,-1{.}32).|

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Use the appropriate method to obtain a one-variable equation
Use the substitution method, since |x^2| in the equation of the parabola is already isolated.||x24−y29=1−4(y−2.75)4−y29=1|| -
Manipulate the equation so that it equals |0|
||−4y+114−y29=19(−4y+11)36−4y236=1−36y+99−4y236=1−36y+99−4y2=364y2+36y−63=0|| -
Solve the equation
Use the quadratic formula to obtain: ||y=−b±√b2−4ac2a=−(36)±√(36)2−4(4)(−63)2(4)=−36±√2 3048y∈{−10.5,1.5}|| -
Substitute the values obtained in one of the starting equations
For |y=-10{.}5|:||x2=−4(y−2.75)=−4(−10.5−2.75)=53x=±√53xA≈−7.28xB≈7.28||For |y=1{.}5|:||x2=−4(y−2.75)=−4(1.5−2.75)=5x=±√5xC≈−2.24xD≈2.24|| -
Write the coordinates of the points of intersection
The coordinates of the four points of intersection between the parabola |x^2=-4(y-2{.}75)| and the hyperbola |\dfrac{x^2}{4}-\dfrac{y^2}{9}=1| are |A(-7{.}28,-10{.}5),| |B(7{.}28,-10{.}5),| |C(-2{.}24,1{.}5),| and |D(2{.}24,1{.}5).|

Determine the coordinates of the point(s) of intersection between the parabolas |(y+4)^2=8(x-2)| and |(y-1)^2=-16(x-10).|
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Use the appropriate method to obtain a one-variable equation
In general, the methods of comparison or substitution are the most efficient for obtaining a one-variable equation. However, we notice that |8| is a factor of |-16.| Therefore, instead of isolating a variable, eliminate the terms in |x| using the elimination method by multiplying the equation |(y+4)^2=8(x-2)| by |-2|.Expand the two equations:||(y+4)2=8(x−2)y2+8y+16=8x−16−2(y2+8y+16)=−2(8x−16)−2y2−16y−32=−16x+32(y−1)2=−16(x−10)y2−2y+1=−16x+160
||Subtracting the two equations gives:||−2y2−16y−32=−16x+ 32−(y2− 2y+ 1=−16x+160)−3y2−14y−33= 0x−128|| -
Manipulate the equation so that it equals |0|
||−3y2−14y−33=−1283y2+14y−95=0|| -
Solve the equation
Use the quadratic formula to obtain: ||y=−b±√b2−4ac2a=−(14)±√(14)2−4(3)(−95)2(3)=−14±√1 3366y∈{−8.43,3.76}|| -
Substitute the values obtained in one of the starting equations
Start by isolating |x| in one of the equations.||(y+4)2=8(x−2)(y+4)28=x−2(y+4)28+2=x||For |y=-8{.}43|:||xA=(y+4)28+2=(−8.43+4)28+2≈4.45||For |y=3{.}76|:||xB=(y+4)28+2=(3.76+4)28+2≈9.53|| -
Write the coordinates of the points of intersection
The coordinates of the two points of intersection between parabolas |(y+4)^2=8(x-2)| and |(y-1)^2=-16(x-10)| are |A(4{.}45,-8{.}43)| and |B(9{.}53,3{.}76).|
