Question

HELP!! The tangent to the curve y^2-xy+9=0 is vertical when??

Answer 1

I get stuck at the part y^2-xy=9. Whats next?

Answer 2

it's \[y^2 - xy + 9 = 0\] or \[y^2 - xy - 9 = 0\] ?

Answer 3

Anyway, let the equation be \[y^2 - xy + 9 = 0\:\:\:(1)\] then \[2yy'(x) - y - xy'(x) = 0\] \[y'(x)= y / (2y -x)\] if you want now to find the tangent you want the derivative to be \[\pm \infty\] so say that \[2y - x = 0 => y = x/2\:\:\:(2)\] for \[y \neq0\] Now combine (1) and (2) and you have \[x^2/4 - x^2/2 + 9 =0 =>...=>x =\pm6\]
So the vertical tangents of (1) are two, \[x=6\] and \[x = -6\]

Answer 4

check the .png file for a visual approach