at the point (-8,6) on the circle given by
.
Tangent and Normal lines: Find the equations of the tangent and
normal lines
at the point (-8,6) on the circle given by .
We can use the fact that the tangent line has a slope equal to the derivative evaluated at the given point.
We can use implicit differentiation to obtain the derivative: Differentiating each term gives
, or 
.
At (-8,6) the derivative becomes 8/6. We can use the
slope intercept form of the line at this point to get 
.
We can substitute the coordinates of the point in order to find the intercept, b1 :
becomes 
,
or 
Since the normal is perpendicular
to the tangent, its slope is the negative reciprocal of the slope of the
tangent line, or -6/8. The equation of the normal becomes 
.
As before, the line goes through the given point, so we can use the coordinates of the point to get the intercept of the normal line, b2 :
or 
,
or 
.
Hence the equation of the normal is 