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Mathematics LibreTexts

2.3: Curve Intersection

  • Page ID
    230
  • [ "article:topic", "authorname:green" ]

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    Intersection of Lines

    Recall that if we want to find the intersection point of two lines, we have two choices:  substitution and elimination.

    Example 1: Substitution

    Solve

    \[ x + 2y = 5,\]

    \[4x - 3y = -2.\]

    Solutions

    We use the first equation to solve for \(x\):

      \[ x = 5 - 2y\]

    then we plug  this into the second equation to get

    \[\begin{align} 4(5 - 2y) - 3y &= -2 \\ -11y + 20 &= -2 \\ y &= 2 \end{align}\]

    and stick this back into the equation for \(x\) to get:

    \[x = 5 - 2(2) = 1.\]

    Example 2: Elimination

    Solve 

    \[ 2x + 5y = 19\]

    \[ 3x - 5y = -9.\]

    We add the equation to get

      \[5x = 10,\; x = 2\]

    Hence

    \[ 2(2) + 5y = 19\]

    \[ y = 3.\]

    Intersection of Other Curves

    Example 3: Substitution

    Find the intersection of the curves

    \[x^2 + y^2 = 25 \]

    and

    \[ y = \dfrac{1}{3}x + 3.\]

    We use the method of substitution to arrive at

    \[\begin{align} x^2+\left(\dfrac{1}{3}x+3\right)^2&=25 \\ x^2+\dfrac{1}{9}x^2+2x+9&=25 \\ 10x^2+18x-144&=0 \\ 5x^2+9x-72&=0 \\ (5x+24)(x-3)&=0 \end{align}\]

    \[ x=-\dfrac{24}{5} \; \text{ or }\; x=3\]

    \[y=(\dfrac{1}{3}) \left(\dfrac{-24}{5}\right)+3 \;\;\; \text{or} \;\;\; y=\dfrac{1}{3}(3)+3\]

    \[y=-\dfrac{7}{5} \; \text{ or } \; y=4. \]

    We get the points 

            \(\left(-\dfrac{24}{5},-\dfrac{7}{5}\right)\) and \((3,4)\).

    Example 4: Elimination

    \[x^2 + 2y^2 = 18\]

    \[ 2x^2 + y^2 = 15\]

    We multiply the first equation by 2 and subtract the second equation to get:

    \[\begin{align} 3y^2&= 21 \\  y^2&= 7 \\ y &= \sqrt{7} \;\;\; \text{or} \;\;\; y = -\sqrt{7} \end{align}\]

    Substituting back into the first equation, we get:

    \[x^2 + 2(7) = 18\]

    \[x = 2 \;\;\; \text{or} \;\;\; x = -2,\]

    hence we get the four points:

       \[ (2,\sqrt{7}), (-2,\sqrt{7}), (2,-\sqrt{7}), (-2,-\sqrt{7}).\]

    Example 5: Using a Graphing Calculator

    We will use a graphing calculator to find the intersection of

    \[y^2+ 16x = 0\]

    and

    \[y^2+ 9x^2-18x = 18.\]

    To find the intersection we just use the intersection function on the graphing calculator.

    Example 6

    We will use the intercept method to solve

    \[(x - 7)(x + 4) = (x + 1)^2.\]

    We find the intersection of the two curves

    \[y = (x - 7)(x + 4)\]

    and 

    \[y = (x + 1)^2.\]

    Larry Green (Lake Tahoe Community College)