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Let $(x_0,y_0)$, where $y_o \neq 0$, be a point on the parabola $y^2=2px$. Find an equation of the tangent line to the parabola passing through $(x_0,y_0)$.

I did my work for this, but I cannot get the answer, which my tutor swears on his job (Joking, but I'm pretty sure its correct), that is correct.

Given Answer: $y_0y=p(x+x_0)$

Here's my workings.

Differentiate with respect to $x$, we have $2y\frac{\mathrm{dy} }{\mathrm{d} x}=2p$ Therefore, $\frac{\mathrm{dy} }{\mathrm{d} x}=\frac {p} {y}$

At Point $(x_0,y_0), \frac{\mathrm{dy} }{\mathrm{d} x}=\frac {p} {y_0}$

Using the equation for straight lines, we have $(y-y_0)=\frac{p}{y_0}(x-x_0)$ $$yy_0-y_0^2=px-px_0$$

By now, it should be clear I will not get the answer. Did I make any mistake?

I reverse-engineered my tutor's answer and I realised he used $(y-y_0)=\frac{p}{y}(x-x_0)$ instead. $y^2-yy_0=px-px_0$




Why is it that when I use $\frac {p}{y}$ versus $\frac {p}{y_0}$, both gives me the same answer? Which should I be using?

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The answer given is quadratic in $x$. This cannot be a line and seems to be incorrect. – ivan Sep 28 '12 at 14:11
Whoops. Edited. there shouldn't be an x – Yellow Skies Sep 28 '12 at 14:16
Just use the fact that $y^2=2px$ for $y_0$ and plug back in. You are already there. – ivan Sep 28 '12 at 14:18
up vote 3 down vote accepted

Since $y_0^2=2px_0$, your solution is equivalent to $yy_0=p(x+x_0)$.

Your "reverse-engineered" solution works only by accident; it has two errors that happen to cancel each other out exactly:

  1. Using $p/y$ as the slope is wrong; that amounts to saying: "to find out whether some point is on the tangent line, find out what the slope of the parabola at that point would have been (if a woodchuck could chuck wood), and check whether the connecting line between the test point and $(x_0,y_0)$ is right". That has no right to work -- the slope you want to compare to is the slope at $(x_0,y_0)$, not the slope at $(x,y)$, no matter whether $(x,y)$ is even on the parabola.

  2. You're rewriting $y^2$ to $2px$, but that is only true when $(x,y)$ is actually on the parabola. This is true by assumption about $(x_0,y_0)$, but the rewriting is supposed to preserve the truth or falsity of the original equation for all $(x,y)$, not just those on the parabola.

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Ok thanks. Didn't see that lol. – Yellow Skies Sep 28 '12 at 14:23
Oh yeah just wondering, I added a couple of steps to my answer. Why is it that when I use $\frac {p}{y}$ versus $\frac {p}{y_0}$, both gives me the same answer? Which should I be using? – Yellow Skies Sep 28 '12 at 14:24
See updated answer. – Henning Makholm Sep 28 '12 at 14:38

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