Question

find any points on the curve x^2+xy=2y-8 where the line tangent to the curve is perpednicular to 2x+4y-1001=0?

Answer 1

first, find the equation of the line perpendicular to 2x+4y-1001=0

Answer 2

ok then what do i do?

Answer 3

find the derivative of y wrt x

Answer 4

\[x^2+xy=2y-8 \]
\[2x+xy'+y=2y'\] solve for \(y'\)

Answer 5

\[2x+4y-1001=0\] has slope \(-\frac{1}{2}\) perpendicular line will have slope \(2\)
set \(y'=2\) and solve

Answer 6

Ohh ok then Thanks!