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I'm not sure whether the definition of a parametric normal line I found online is right. I'm trying to answer problem 41. I'll use their formula. The gradient at the point (-3,4) is <-6,8>, so the equation of the normal line has $ x(t)=-3+-6t$ and $ y(t)=4+8t$ . Side question: how do I convert this parametric equation to cartesian equation/coordinates?

Have a successfully found a parametric equation for the line that is perpendicular to the graph of the given equation at the given point?

We can treat the circle as a level curve and we know that the gradient is always perpendicular to the level curves.

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Yes, you've found a normal line, try drawing a picture and you'll be sure of it!

Hints for converting to a function $y=f(x)$:

1. We have this system of equations:

$x(t) = -3 - 6t$

$y(t) = 4+8t$

What is $x+y$?

2. We can write:

$[x(t),y(t)]=[-3,4]+t[-6,8]$

From this, you might be able to figure out what the slope of our function should be.

Good luck!

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