given : $x^2+6x+2y-8=0$ at $x = 3$
$y = \frac{17}{2}$
$y' = 0 = m_T$
I was able to find the equation of the tangent line which is $y=\frac{17}{2}$ by using the point-slope formula however, when finding $m_N$ (slope of the normal line), I would get an indeterminate.
$m_T ( m_N )= -1$
$m_N = -\frac{1}{0} $
thus, gives me : $y - \frac{17}{2} = -\frac{1}{0}(x-3)$
how do I solve this equation?
I know that the equation of the normal line is $x=3$, which I found by graphing, however I just wanted to know if I can show my solution through an equation.
am I not breaking any rules by doing this? $-0(y-\frac{17}{2})=1(x-3)$
$\endgroup$3 Answers
$\begingroup$HINT
The equation for a vertical line is in the form $$x=k$$
More in general, to avoid indeterminate form, note that the line equation in implicit form is $ax+by=c$ and the condition for the line to be normal with the line $dx+ey=f$ is given by
$$ad+be=0$$
$\endgroup$ $\begingroup$Recall that the normal line must be perpendicular to the tangent line and they both meet at the point of your curve (namely where $x=3$ in this particular case). In this situation, your tangent line is horizontal and therefore the normal line must be a vertical line since it has to be perpendicular to your tangent line. Vertical lines have the form $$x=k$$ for some $k$. In your case, the normal line must pass by the point $\left(3,\frac{17}2\right),$ thus have equation $x=3$.
The method that you outline above works for most situations. In fact, it works as long as your tangent line is not vertical or horizontal. The reason why it fails in that situation is that the normal line is not a function of $y$ in terms of $x$. As you can see from the graph, for $x=3$, your normal line has infinitely $y$ values. But this will be the case whenever your derivative at a point is $0$. In that case, you know that the normal line must be of the form $x=k$ by the argument above.
$\endgroup$ $\begingroup$It looks there is an error in calculation. The function is $$y=-\frac12x^2-3x+4, y(3)=-9.5.$$ The slope of tangent at $x=3$ is $$m_T=y'(3)=-3-3=-6.$$ The slope of normal (perpendicular to tangent) is $$m_N=-\frac{1}{m_T}=\frac16.$$ The equation of normal passing through $(3,-9.5)$ is $$-9.5=\frac{1}{6}\cdot 3+b \Rightarrow b=-10 \Rightarrow y_N=\frac16x-10.$$
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