Answer:
C) One vertical but no horizontal
Step-by-step explanation:
Use implicit differentiation to find dy/dx.
xy² = 2 + xy
x (2y dy/dx) + y² = x dy/dx + y
(2xy − x) dy/dx = y − y²
dy/dx = (y − y²) / (2xy − x)
When the tangent lines are horizontal, dy/dx = 0.
0 = (y − y²) / (2xy − x)
0 = y − y²
0 = y (1 − y)
y = 0 or 1
When the tangent lines are vertical, dy/dx is undefined, so the denominator is 0.
0 = 2xy − x
0 = x (2y − 1)
x = 0, y = 1/2
There are two possible points where there can be a horizontal tangent line, and two possible points where there can be a vertical tangent line. Plug all back into the original equation to see which are actual points.
If y = 0:
x(0)² = 2 + x(0)
0 = 2
No solution.
If y = 1:
x(1)² = 2 + x(1)
x = 2 + x
0 = 2
No solution.
If x = 0:
(0)y² = 2 + (0)y
0 = 2
No solution.
If y = 1/2:
x(1/2)² = 2 + x(1/2)
x/4 = 2 + x/2
-x/4 = 2
x = -8
Therefore, there is one vertical line and no horizontal lines.
