Implicit Differentiation: hor., vert. tangents to xy = 16y^2

[Becky4paws]

Find all points of the given curve where the tangent line is (a) horizontal and (b) vertical.

xy = 16y^2

x'y + xy' dy/dx = 32y dy/dx
xy' - 32y dy/dx = -x'y
dy/dx = -x'y/y(x-32)

Assuming that is correct, I thought in order to find horizontal and vertical points, you evaluated the numerator = 0 (horizontal) and denominator = 0 (vertical).

a. -x'y = 0
-x' = 0/y

If that is the case, it doesn't exist.

b. y(x-32) = 0
y = 0, x = 32

Vertical point at (32.0)

I hope I'm not too far off.....

 

[galactus]

Differentiate implicitly wrt y:

\(\displaystyle \L\\xy-16y^{2}=0\)

\(\displaystyle \L\\x\cdot\frac{dy}{dx}+y-32y\frac{dy}{dx}=0\)

\(\displaystyle \L\\\frac{dy}{dx}=\frac{-y}{x-32y}\)


Now, wrt to x:

\(\displaystyle \L\\x+y\frac{dx}{dy}-32y=0\)

\(\displaystyle \L\\\frac{dx}{dy}=\frac{32y-x}{y}\)

If you think about it, what does your function represent?. That may tell you something of the slope.