What is the slope of the tangent line of $(x - 2) (y - 3) - e^{y} = C$ $(x-2)(y-3)-e^y= C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?

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Answer 1 · 2017-01-16T20:53:11

$y'|_{(-2,1)} = - (2)/(e +4)$

We can use implicit differentiation (and the product rule for the first term):

$(y-3) + (x-2) y' - y' e^y= 0$

$(y-3) = y' (e^y - x + 2)$

$y' = (y-3)/(e^y - x + 2)$

$y'|_{(-2,1)} = (1-3)/(e^1 - (-2) + 2) = - (2)/(e +4)$

we can also check this by evaluating the normal vector $mathbf n$ as

$mathbf n = nabla ((x-2)(y-3)-e^y) = langle y-3, x-2 - e^y rangle$

$mathbf n_{(-2,1)} = langle -2, -4 - e rangle$

and for tangent vector $mathbf t$ , we say:

$mathbf n * mathbf t = 0 implies \mathbf t = langle 4 + e, -2, rangle$ , same answer

What is the slope of the tangent line of (x-2)(y-3)-e^y= C (x−2)(y−3)−ey=C (x-2)(y-3)-e^y= C , where C is an arbitrary constant, at (-2,1)(−2,1)(-2,1)?