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Sorry but I'm a bit stuck at this problem with derivatives and the tangent line, I'll be stating the problem then the steps I've done to try and solve it. The problem is stated like this:

Find all points on the graph of $y = (x-2)^2$ at which the tangent line is perpendicular to the line with equation $2x - y + 2 = 0$

So I have the equation for the normal line: $y = 2(x + 1)$, so I think the equation for the tangent line is $y = -\frac{1}{2}(x + 1)$.

Since I need to get points of intersections with $(x-2)^2$, I need to get an equation for the tangent line based on it so I solve for derivatives.

$$f(x) = (x - 2)^2$$ $$f'(x) = \frac{d}{dx} (x-2)^2$$ $$f'(x) = 2(x - 2)$$

So now given these, I have another equation for the tangent line at a given point of tangency $t$:

$$y - f(t) = f'(t)(x - t)$$ $$y - (t - 2)^2 = 2(t - 2)(x - t)$$ $$y = 2(t - 2)(x - t) + (t - 2)^2$$ $$y = (t - 2)[2(x - t) + (t - 2)]$$ $$y = (t - 2)(2x - 2t + t - 2)$$ $$y = (t - 2)(2x -t - 2)$$ $$y = (t - 2)2(x - \frac{t}{2} - 1)$$ $$y = 2(t - 2)(x - \frac{t + 2}{2})$$

Now going back to the equation for the normal line, I now have two equations for the tangent line:

$$y = -\frac{1}{2}(x + 1)$$ $$y = 2(t - 2)(x - \frac{t + 2}{2})$$

Now this is where I'm stuck. The two equations seem to imply that $2(t - 2) = -\frac{1}{2}$ and $\frac{t + 2}{2} = -1$, however solving for $t$ on the first equation gives me $\frac{7}{4}$ which seem to match what's in the answer key, but the other equation gives me $t = -4$. Why is that?

Was I wrong in any of my assumptions? I know I must be missing something huge but I can't pinpoint it right now. Any help would be greatly appreciated.

Thanks.

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    $\begingroup$ Because you only need to compare the slope, and not the point of intersection with x-axis $\endgroup$
    – hjpotter92
    Sep 12, 2016 at 10:12

2 Answers 2

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the slope of the equation $$y=(x-2)^2$$ is given by $$y'=2(x_0-2)$$ the slope of the erquation $$2x-y+2=0$$ is given by $$m_2=2$$ now it must be $$m_1\cdot m_2=-1$$ and you will get $$2\cdot 2(x_0-2)=-1$$ can you proceed?

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  • $\begingroup$ Yes! I can't believe I went through all the trouble of solving for those equations :/ Our texts usually say that for a given $f$, we should be able to get the normal line equation if we have the tangent line equation at point $t$ with the slope $f'(t)$. For some reason I thought that I could do the inverse (define a tangent line equation given a normal line equation) but without the point of tangency. Thank you. I'll accept as soon as I'm able $\endgroup$
    – R. Cruz
    Sep 12, 2016 at 10:21
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There are an infinite number of perpendicular lines through a given line. If the normal line has equation $y = 2x + 2$, then the tangent line has equation $y = -\frac{1}{2} x + q$ for some $q \in \mathbb{R}$.

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  • $\begingroup$ Yes, thanks for pointing it out. I was a bit naive in thinking that I could get the equation for the tangent line without a definite point of intersection. $\endgroup$
    – R. Cruz
    Sep 12, 2016 at 10:23

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