Equations of All Tangents
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Calc12 said:
Write the equation of a line that has a given slope (df/dx) and passes through a given point (0,0).
Hello, Calc12!
Let the equation of the tangent be: .\(\displaystyle y \:=\:mx\)
If this line is tangent to the parabola, there will be one point of intersection.
\(\displaystyle \text{We have }\,(x^2\;-\;4x\;+\;25) \,\cap\,(mx) \quad\Rightarrow\quad x^2\;-\;4x\;+\;25 \:=\:mx \quad\Rightarrow\quad x^2 \;-\; (m+4)x \;+\; 25 \:=\:0\)
. . \(\displaystyle \text{Quadratic Formula: }\;x \;=\;\frac{(m+4) \pm\sqrt{(m+4)^2 - 4(1)(25)}}{2(1)} \;=\; \frac{(m+4) \pm\sqrt{m^2+8m - 84}}{2}\)
\(\displaystyle \text{If the quadratic equation has one root, the discriminant must be zero: }\;m^2 \;+\; 8m \;-\; 84 \:=\:0\)
. . \(\displaystyle \text{We have: }\
m - 6)(m + 14) \:=\:0 \quad\Rightarrow\quad m \:=\:6,\:-14\)
\(\displaystyle \text{Therefore, the equations of the tangents are: }\;\begin{Bmatrix} y &=& 6x \\ \\[-3mm] y &=& \text{-}14x \end{Bmatrix}\)
Let the equation of the tangent be: .\(\displaystyle y \:=\:mx\)
If this line is tangent to the parabola, there will be one point of intersection.
\(\displaystyle \text{We have }\,(x^2\;-\;4x\;+\;25) \,\cap\,(mx) \quad\Rightarrow\quad x^2\;-\;4x\;+\;25 \:=\:mx \quad\Rightarrow\quad x^2 \;-\; (m+4)x \;+\; 25 \:=\:0\)
. . \(\displaystyle \text{Quadratic Formula: }\;x \;=\;\frac{(m+4) \pm\sqrt{(m+4)^2 - 4(1)(25)}}{2(1)} \;=\; \frac{(m+4) \pm\sqrt{m^2+8m - 84}}{2}\)
\(\displaystyle \text{If the quadratic equation has one root, the discriminant must be zero: }\;m^2 \;+\; 8m \;-\; 84 \:=\:0\)
. . \(\displaystyle \text{We have: }\
m - 6)(m + 14) \:=\:0 \quad\Rightarrow\quad m \:=\:6,\:-14\)\(\displaystyle \text{Therefore, the equations of the tangents are: }\;\begin{Bmatrix} y &=& 6x \\ \\[-3mm] y &=& \text{-}14x \end{Bmatrix}\)
Let (p,q) be the point on the parabola where the line is tangent.
The slope at any point is then \(\displaystyle m=2p-4\)
and \(\displaystyle q=p^{2}-4p+25\)
Both lines pass through the origin, so \(\displaystyle p_{1}=0, \;\ q_{1}=0\)
Now, use \(\displaystyle q-q_{1}=m(p-p_{1})\) and solve for p.
This gives the points on the parabola where the lines are tangent.
The line equations can easily be found by having (0,0) and the two points of tangency.
The slope at any point is then \(\displaystyle m=2p-4\)
and \(\displaystyle q=p^{2}-4p+25\)
Both lines pass through the origin, so \(\displaystyle p_{1}=0, \;\ q_{1}=0\)
Now, use \(\displaystyle q-q_{1}=m(p-p_{1})\) and solve for p.
This gives the points on the parabola where the lines are tangent.
The line equations can easily be found by having (0,0) and the two points of tangency.