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? $-0(y-\frac{17}{2})=1(x-3)$
Best Answer
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
$$ad+be=0$$