Points of Intersection Between a Parabola and a Conic

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m1333

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points-of-intersection-between-a-parabola-and-a-conic

Parent content

Conics

Grades

Secondary V

Topic

Mathematics

Tags

conic

circle

ellipse

hyperbola

parabola

points of intersection

meeting points

crossing points

intersection between a parabola and a circle

intersection between a parabola and an ellipse

intersection between a parabola and a hyperbola

intersection between two parabolas

Content

Contenu

Corps

To find the intersection point(s) between a parabola and a conic, we solve one or more system(s) of second-degree equations.

Content

Corps

Use the appropriate method (comparison, substitution, or elimination) to obtain an equation with one variable.
Manipulate the equation so that it equals |0.|
Solve the equation to find the value(s) of the isolated variable.
Replace the value(s) obtained in one of the original equations to obtain the value(s) of the other variable.
Write the coordinates of the point(s) of intersection.

Content

Corps

Unlike the intersection between a line and a conic, there are five possible cases regarding the number of solutions:

the parabola and the conic do not intersect;
the parabola and the conic intersect only at one place, called the point of tangency; or
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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https://www.geogebra.org/material/iframe/id/qf6ydnjx/width/600/height/600/borde…

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600

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600

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Type

Anchor

Title

The Intersection Points Between a Parabola and a Circle

Anchor

parabola-circle

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Anchor

Title

The Intersection Points Between a Parabola and an Ellipse

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parabola-ellipse

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Anchor

Title

The Intersection Points Between a Parabola and a Hyperbola

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parabola-hyperbola

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Anchor

Title

The Intersection Points Between Two Parabolas

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two-parabolas

Title (level 2)

The Intersection Points Between a Parabola and a Circle

Title slug (identifier)

parabola-circle

Contenu

Content

Corps

Determine the coordinates of the point(s) of intersection between the parabola |y^2=-16x| and the circle |x^2+y^2=36.|

Solution

Corps

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.||
$\begin{align}x^2+\color{#3a9a38}{y^2}&=36\\x^2+\color{#3a9a38}{-16x}&=36\end{align}$ ||
Manipulate the equation so that it equals |0|
||
$\begin{align}x^2-16x&=36\\ \color{#3B87CD}{1}x^2\color{#3A9A38}{-16}x\color{#EC0000}{-36}&=0\end{align}$ ||
Solve the equation
Use the quadratic formula to obtain:||
$\begin{align} x&=\dfrac{-\color{#3A9A38}{b}\pm\sqrt{\color{#3A9A38}{b}^2-4\color{#3B87CD}{a}\color{#EC0000}{c}}}{2\color{#3B87CD}{a}}\\\\ &=\dfrac{-(\color{#3A9A38}{-16})\pm\sqrt{(\color{#3A9A38}{-16})^2-4(\color{#3B87CD}{1})(\color{#EC0000}{-36})}}{2(\color{#3B87CD}{1})}\\\\ &=\dfrac{16\pm\sqrt{400}}{2}\\\\ x&\in\{-2,18\} \end{align}$ ||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
||
$\begin{align} y^2&=-16\color{#3A9A38}{x}\\ &=-16(\color{#3A9A38}{-2})\\ &=32\\ y&=\pm\sqrt{32}\\\\ y_{\small{A}}&\approx5{.}66\\ y_{\small{B}}&\approx-5{.}66 \end{align}$ ||
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).|

Image

A parabola intersecting a circle at two distinct points.

Title (level 2)

The Intersection Points Between a Parabola and an Ellipse

Title slug (identifier)

parabola-ellipse

Contenu

Content

Corps

Determine the coordinates of the point(s) of intersection between the parabola |x^2=12(y+3)| and the ellipse |\dfrac{x^2}{25}+\dfrac{y^2}{9}=1.|

Solution

Corps

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.||
$\begin{align} \dfrac{\color{#3a9a38}{x^2}}{25}+\dfrac{y^2}{9}&=1\\\\ \dfrac{\color{#3a9a38}{12(y+3)}}{25}+\dfrac{y^2}{9}&=1\end{align}$ ||
Manipulate the equation so that it equals |0|
||
$\begin{align}\dfrac{12y+36}{25}+\dfrac{y^2}{9}&=1\\\\ \dfrac{9(12y+36)}{225}+\dfrac{25y^2}{225}&=1\\\\ \dfrac{108y+324+25y^2}{225}&=1\\\\ 25y^2+108y+324&=225\\ \color{#3B87CD}{25}y^2+\color{#3A9A38}{108}y+\color{#EC0000}{99}&=0\end{align}$ ||
Solve the equation
Use the quadratic formula to obtain: ||
$\begin{align} y&=\dfrac{-\color{#3A9A38}{b}\pm\sqrt{\color{#3A9A38}{b}^2-4\color{#3B87CD}{a}\color{#EC0000}{c}}}{2\color{#3B87CD}{a}}\\\\ &=\dfrac{-(\color{#3A9A38}{108})\pm\sqrt{(\color{#3A9A38}{108})^2-4(\color{#3B87CD}{25})(\color{#EC0000}{99})}}{2(\color{#3B87CD}{25})}\\\\ &=\dfrac{-108\pm\sqrt{1\ 764}}{50}\\\\ y&\in\{-3,-1{.}32\} \end{align}$ ||
Substitute the values obtained in one of the starting equations
For |y=-3|: ||
$\begin{align} x^2&=12(\color{#3a9a38}{y}+3)\\ &=12(\color{#3a9a38}{-3}+3)\\ &=0\\\\ x_{\small{A}}&=0 \end{align}$ ||For |y=-1{.}32|:|| $\begin{align} x^2&=12(\color{#3a9a38}{y}+3)\\ &=12(\color{#3a9a38}{-1{.}32}+3)\\ &=20{.}16\\ x&=\pm\sqrt{20{.}16}\\\\ x_{\small{B}}&\approx-4{.}49\\ x_{\small{C}}&\approx4{.}49 \end{align}$ ||
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).|

Image

A parabola intersecting an ellipse at three distinct points.

Title (level 2)

The Intersection Points Between a Parabola and a Hyperbola

Title slug (identifier)

parabola-hyperbola

Contenu

Content

Corps

Determine the coordinates of the point(s) of intersection between the parabola |x^2=-4(y-2{.}75)| and hyperbola |\dfrac{x^2}{4}-\dfrac{y^2}{9}=1.|

Solution

Corps

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.||
$\begin{align} \dfrac{\color{#3a9a38}{x^2}}{4}-\dfrac{y^2}{9}&=1\\\\ \dfrac{\color{#3a9a38}{-4(y-2{.}75)}}{4}-\dfrac{y^2}{9}&=1\end{align}$ ||
Manipulate the equation so that it equals |0|
||
$\begin{align}\dfrac{-4y+11}{4}-\dfrac{y^2}{9}&=1\\\\ \dfrac{9(-4y+11)}{36}-\dfrac{4y^2}{36}&=1\\\\ \dfrac{-36y+99-4y^2}{36}&=1\\\\ -36y+99-4y^2&=36\\ \color{#3B87CD}{4}y^2+\color{#3A9A38}{36}y\color{#EC0000}{-63}&=0\end{align}$ ||
Solve the equation
Use the quadratic formula to obtain: ||
$\begin{align} y&=\dfrac{-\color{#3A9A38}{b}\pm\sqrt{\color{#3A9A38}{b}^2-4\color{#3B87CD}{a}\color{#EC0000}{c}}}{2\color{#3B87CD}{a}}\\\\ &=\dfrac{-(\color{#3A9A38}{36})\pm\sqrt{(\color{#3A9A38}{36})^2-4(\color{#3B87CD}{4})(\color{#EC0000}{-63})}}{2(\color{#3B87CD}{4})}\\\\ &=\dfrac{-36\pm\sqrt{2\ 304}}{8}\\\\ y&\in\{-10{.}5,1{.}5\} \end{align}$ ||
Substitute the values obtained in one of the starting equations
For |y=-10{.}5|:||
$\begin{align} x^2&=-4(\color{#3a9a38}{y}-2{.}75)\\ &=-4(\color{#3a9a38}{-10{.}5}-2{.}75)\\ &=53\\ x&=\pm\sqrt{53}\\\\ x_{\small{A}}&\approx-7{.}28\\ x_{\small{B}}&\approx7{.}28 \end{align}$ ||For |y=1{.}5|:|| $\begin{align} x^2&=-4(\color{#3a9a38}{y}-2{.}75)\\ &=-4(\color{#3a9a38}{1{.}5}-2{.}75)\\ &=5\\ x&=\pm\sqrt{5}\\\\ x_{\small{C}}&\approx-2{.}24\\ x_{\small{D}}&\approx2{.}24 \end{align}$ ||
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).|

Image

A parabola intersecting a hyperbola at four distinct points.

Title (level 2)

The Intersection Points Between Two Parabolas

Title slug (identifier)

two-parabolas

Contenu

Content

Corps

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).|

Solution

Corps

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:||
$\begin{align} (y+4)^2&=8(x-2)\\ y^2+8y+16&=8x-16\\ \color{#EC0000}{-2}(y^2+8y+16)&=\color{#EC0000}{-2}(8x-16)\\ -2y^2-16y-32&=\color{#3B87CD}{-16}x+32\\\\ (y-1)^2&=-16(x-10)\\ y^2-2y+1&=\color{#3B87CD}{-16}x+160 \end{align}$ ||Subtracting the two equations gives:|| $\begin{align} -2y^2-16y-32&=-16x+\ \ 32\\ ^{\huge{-}}\quad\quad (y^2-\ \ 2y+\ \ 1&=-16x+160)\\ \hline -3y^2-14y-33&= \ \ \ \ \ \color{#EC0000}{0x}-128 \end{align}$ ||
Manipulate the equation so that it equals |0|
||
$\begin{align}-3y^2-14y-33&=-128\\ \color{#3B87CD}{3}y^2+\color{#3A9A38}{14}y\color{#EC0000}{-95}&=0\end{align}$ ||
Solve the equation
Use the quadratic formula to obtain: ||
$\begin{align} y&=\dfrac{-\color{#3A9A38}{b}\pm\sqrt{\color{#3A9A38}{b}^2-4\color{#3B87CD}{a}\color{#EC0000}{c}}}{2\color{#3B87CD}{a}}\\\\ &=\dfrac{-(\color{#3A9A38}{14})\pm\sqrt{(\color{#3A9A38}{14})^2-4(\color{#3B87CD}{3})(\color{#EC0000}{-95})}}{2(\color{#3B87CD}{3})}\\\\ &=\dfrac{-14\pm\sqrt{1\ 336}}{6}\\\\ y&\in\{-8{.}43,3{.}76\} \end{align}$ ||
Substitute the values obtained in one of the starting equations
Start by isolating |x| in one of the equations.||
$\begin{align}(y+4)^2&=8(x-2)\\ \dfrac{(y+4)^2}{8}&=x-2\\ \dfrac{(y+4)^2}{8}+2&=x\end{align}$ ||For |y=-8{.}43|:|| $\begin{align}x_{\small{A}}&=\dfrac{(\color{#3A9A38}{y}+4)^2}{8}+2\\\\ &=\dfrac{(\color{#3A9A38}{-8{.}43}+4)^2}{8}+2\\\\ &\approx4{.}45\end{align}$ ||For |y=3{.}76|:|| $\begin{align} x_{\small{B}}&=\dfrac{(\color{#3A9A38}{y}+4)^2}{8}+2\\\\ &=\dfrac{(\color{#3A9A38}{3{.}76}+4)^2}{8}+2\\\\ &\approx9{.}53\end{align}$ ||
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).|

Image

A parabola intersecting another parabola at two distinct points.

Title (level 2)