Question

How do you do this question?

Answer 1

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.