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Find the point on the graph of $y=e^{2x}$ at which the tangent line passes through the origin.

Completely lost on this question, the wording is confusing here.

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It's not really a question as it is. Are you supposed to find that point? –  Karolis Juodelė May 10 '13 at 13:16
    
Looks like I made a mistake, the 'for' should say find. –  adeshina lawel May 10 '13 at 13:20
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@Closevoter Why? I don't think this is a "problem statement question"--the OP is asking for help understanding the question, not for someone to do the problem for them. –  anorton May 10 '13 at 13:28

4 Answers 4

up vote 2 down vote accepted

We have a mystery point $P$ on the curve. The tangent line at $P$ goes through the origin. We want to find the coordinates of $P$.

Let the coordinates of $P$ be $(a,e^{2a})$. We find the equation of the tangent line to $y=e^{2x}$ at $P$.

The slope of the tangent line at $P$ is $2e^{2a}$. So the equation of the tangent line is $$y-e^{2a} =2e^{2a}(x-a).$$ This line passes through the origin. So $(0,0)$ is on the line, and therefore $$-e^{2a}=(2e^{2a})(-a).$$ Cancel the $e^{2a}$. We get $a=\frac{1}{2}$.

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Thanks, I got it now. –  adeshina lawel May 10 '13 at 13:34

Some Hints:

You know that a line has the slope $y=mx+b$. To pass through the origin, $b=0$, leaving us with the equation $y=mx$. The derivative of a function gives you the slope of the tangent line, or in the notation I've been using thus far, $m$.

So, this question is asking for a point $(x_0, y_0)$ such that $y_0 = mx_0$.

(Hint: $m$ will actually be a function of $x$ and $y$)

If you need more help, just leave a comment.

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I understand the question now, the wording of the question made me believe that the line was tangent to the graph at the origin. –  adeshina lawel May 10 '13 at 13:35
    
@adeshinalawel Glad to help. :) –  anorton May 10 '13 at 13:37

If $(x_0;y_0)$ is your point, then the straight line passing thru it and tangent to the graph is $$y-e^{2x_0}=2e^{2x_0}(x-x_0).$$ If the origin lies on this line, then $x_0=1/2$.

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Look at the graph of $y = e^{2x}$. For every point on that graph, you can imagine a straight-line that is tangent to it (i.e. goes through that point and has the same slope as the graph at that point). The question's asking you to find the line that goes through the origin; $(0,0)$.

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