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The way to find equation of tangent line of a parabola that has equation $y=Ax^{2}+Bx+C$ and parallel to $Ay=Bx+C$ line is by using $y^{'}=\tfrac{B}{A}$ and some further steps with $y^{'}$ is first-derivative and $\tfrac{B}{A}$ is gradient/slope.

But I've read in my textbook wrote that if a parabola equation is $y^{2}=4px$, the tangent line of parabola equation that parallel to $Ay=Bx+C$ line is $y=mx+\tfrac{p}{m}$ with $m$ is gradient/slope.

Are they have same solutions (to find equation of tangent line) or they have different concepts?

làntèrn
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  • Your method is correct except that it should be $y'=B/A$. The tangent line from the textbook method is apparently not a line. Did you copy it correctly? – KittyL Apr 15 '15 at 17:50
  • Edited the gradient, parabola equation, and the tangent line equation. – làntèrn Apr 15 '15 at 18:14

1 Answers1

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Yes, they are the same. In the textbook method, $2ydy=4pdx\rightarrow \frac{dy}{dx}=\frac{2p}{y}$. The slope $m=\frac{B}{A}=\frac{2p}{y}$. So $y=\frac{2p}{m}$.

Plugging this into the original equation $y^2=4px$ gives you $x=\frac{p}{m^2}$. Using the slope-intercept form of a line $y=mx+b$ would give you $b=\frac{p}{m}$.

KittyL
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