# Finding the equation of the normal line

I have a question to find the equations of the tangent line and the normal line to the curve at the given point. I can find the equation for the tangent line easily but I am not sure what a normal line is and there is no example that I can find.

$y=x^4 + 2e^x$ at (0,2)

From that I do see that if I plug in 0 I get 2 as a result so my guess was that if I plug in another number I can use that to get the slope but it gave an incorrect answer.

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"Normal" in this situation means perpendicular. So instead of finding the equation of the tangent line, you want to find the equation of the "perpendicular line." –  froggie May 9 '12 at 23:59
And what you need to know is that if two lines are perpendicular, the product of their slopes is $-1$. –  Brian M. Scott May 10 '12 at 0:01
This complicates things much more, so basically I still need to find the slope of this equation and then I need to make it multiply by -1? –  user138246 May 10 '12 at 0:03
Not quite: if $ab=-1$, then $a=\frac{-1}b$, not $(-1)b$. –  Brian M. Scott May 10 '12 at 0:07
I am slightly confused, wouldn't I just multiply the curve by -1? –  user138246 May 10 '12 at 0:16

You know that the tangent at the curve is given by

$$y_t = f(a)+f'(a)(x-a)$$

The normal would be a line such that

• It also passes through $(a,f(a))$
• It is perpendicular to $y_t$.

Given that a line that passes through $(X,Y)$ and has slope $m$ is given by

$$y-Y = m(x-X)$$

...and that two lines of slopes $n$ and $p$ are perpendicular if and only if $m\cdot n=-1$

Can you find $y_n$?

Given that a line that passes through $(X,Y)$ and has slope $m$ is given by

$$y-Y = m(x-X)$$

Give the equations to

1. A line with slope $10$ that passes through $(0,1)$
2. A line with slope $-5$ that passes through $(-3,3)$
3. Given a line with slope $2$ that passes through the origin, find the equation to a line perpendicular to it that passes through $(5,2)$.
4. Let $f(x) = x^4+2e^x$. Given the equation of the tangent to $f(x)$ at $(a,f(a))$. Find the normal to $f$ at the same point.
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This post is a little confusing to me. All I did was take the slope and fine the negative inverse of it. Is that wrong? –  user138246 May 10 '12 at 1:53
@Jordan I sense you are lacking some theory. If the point slope definition of a line troubles you, maybe you should go back to the basics for a while and get a hold of that first. I'll give you some excercises, you answer them editing the question. –  Pedro Tamaroff May 10 '12 at 1:56
I understand point slope form well enough I think. –  user138246 May 10 '12 at 2:00
@Jordan Then I don't see what should trouble you. –  Pedro Tamaroff May 10 '12 at 2:08
yn is the line that is the normal line? –  user138246 May 10 '12 at 2:12

Finding the derivative will give you the slope of the tangent line like what Peter T. has shown above. All you need to do is be able to find what (a, f(a)) and then you will be on your way.

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Equation of normal is given by,

$$\frac{y-y_1}{x-x_1}=\frac{-dx}{dy}$$

Now $y=x^4 + 2e^x$

$\frac{dy}{dx}=4x^3+2e^x =2$ .....at $x=0$

$\frac{-dx}{dy}=\frac{-1}{2}$

and $x_1=0$ and $y_1=2$ , plugging these values in the equation for normal, we get

$$\frac{y-2}{x-0}=\frac{-1}{2}$$

$$x+2y=4$$

were you getting the same answer? or something else..

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