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Find the coordinates of the point in the graph of $f(x)=2x^2+3$ that is closer to the point $(5,-1)$

To start off, I found the first order derivative of the function so I could get the slope of the tangent line to $f(x)$

$$f´(x)=4x$$

Then, I needed to find the equation of the perpendicular line that goes through $(5,-1)$.

If $m_t*m_n=-1,$ therefore $m_n=-{1 \over4x}$

And then I proceeded to find the equation of the perpendicular line, getting as a result:

$$y=-\frac 54+ {5 \over 4x} $$

But as you can see, that's not a line. So I'm asking, what am I doing wrong? I'm thinking I have it wrong since I stated the slope of the tangent line is $4x$, I guessing I need a number.

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2 Answers 2

up vote 1 down vote accepted

You need to find a distance between any given point on parabola and $(5, 1)$ and find it's minimum with respect to $x$. So $$d^2(x) = (x - 5)^2 + (2x^2 + 3 + 1)^2 = x^2-10x + 25 + 4(x^4 + 4x^2 + 4) = 4x^4 + 17x^2-10x+41$$ $$\left[ d^2(x) \right]' = 16x^3 + 34x - 10 = 0$$ Latter cubic equation has roots, but it looks like they're not rational, or I messed up somewhere with numbers. But idea holds anyway.

But if you want to stick to your method, you need to:

$ \displaystyle a)\ k = -\frac 1{4x_0}\\ \displaystyle b)\ -1 = -\frac 5{4x_0} + b \quad \Rightarrow \quad b = -1 + \frac 5{4x_0} \\ \displaystyle c)\ 2x_0^2+3=-\frac 14 - 1 + \frac 5{4x_0} \\ \displaystyle d)\ 8x_0^2 + 17x_0-5=0 $

which is essentially the same equation.

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It looks fine to me, but then again, it would be hectic to solve that cubic equation on paper. –  ChairOTP Nov 14 '12 at 23:49
    
Well, all I can say if you want to use normal to the tangent line that goes through your point $(5, 2)$ you'd end up with cubic equation anyway. –  Kaster Nov 14 '12 at 23:55
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Rather than use the tangent, you can just take a point on the graph and calculate the distance to $(5,-1)$. An arbitrary point on the curve is $(x,2x^2+3)$ and the distance is $\sqrt {(x-5)^2 + (2x^2+4)^2}$. We might as well minimize the square of the distance. The derivative of the square of the distance is $2(x-5)+8x(2x^2+4)$. Setting this to zero gives $16x^3+34x-10=0,$ which doesn't have nice roots.

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does that mean that I have to solve it this way and not the way I was doing it, because my teacher did it like that in class, and asked us to do it the other method as homework. –  ChairOTP Nov 14 '12 at 23:31
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@ChairOTP: you can do it the way you tried, but you used $x$ two different ways. Once it is the point on the curve and once it is the variable in the line from the point on the curve to $(5,-1)$. If $a$ is the $x$ coordinate of the point on the curve, the slope of the tangent is $4a$, the slope of the normal is $-\frac 1{4a}$ and the equation line through $(5,-1)$ and parallel to the normal is $y+1=-\frac 1{4a}(x-5)$. Then choose $a$ so this hits $(a, 2a^2+3)$ –  Ross Millikan Nov 14 '12 at 23:37
    
@RossMillikan you forgot a square for $2x^2 + 3$ when found your distance and didn't substract $-1$. –  Kaster Nov 14 '12 at 23:38
    
@Kaster: and it propagated. Fixed now. Thanks –  Ross Millikan Nov 14 '12 at 23:42
    
@RossMillikan May I ask how did you solve for x in the cubic equation? I only know either synthetic division or using a program, and for this one I think the latter one is likely to be used. –  ChairOTP Nov 14 '12 at 23:46
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