Need help in calculus....?

Question

the tangent to the curve y^2 - xy + 9 = 0 is vertical when.....


so i take the derivative and i get 2ydy/dx - y - xdy/dx = 0. then i solve for dy/dx and get y/2y-x..... then im stuck.

Answer 1

Here is a different way to approach it. Let's solve for x to get:

x = y + 9/y

then dx/dy = 1 - 9/y^2

Now for the tangent to be vertical, dx/dy = 0,

which is the case when y=-3,3

which is at points: (-6,-3) and (6,3). Note that your formula verifies that dy/dx is infinite/undefined at these points.

Answer 2

put x= f(y)===>x= (y^2+9)/y = y+ 9/y

dy/dx for a vertical tangent is NOT zero but infinite if you have y=g(x)

so dx/dy should be zero if you take x =f(y)

dx/dy= 1-9/y^2 so y^2 -9 = 0 and y=+-3

the points are (6,3) and(-3,-6)

Answer 3

Ben is correct. The tangent to the curve is vertical when dy/dx is undefined (because of dividing by zero) OR when dx/dy = 0.

Answer 4

This equation is variable separable:
xy = y^2 + 9
x = (y^2 + 9)/y

dx/dy = (2y^2 - y^2 - 9)/y^2
dx/dy = (y^2 - 9)/y^2 = 0 when the tangent is vertical.
y = 3
x = (9 +9)/3
x = 6,
so, at the point (6,3) the tangent to the curve is vertical

y = (x ± √(x^2 - 36))/2