# Help with a math differentiation qn please?

For the curve with equation x^2 - y^2 + 2xy + 4 = 0, find the coordinates of each point at which the tangent is parallel to the x axisl

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- MechEng2030Lv 78 years agoFavorite Answer
Differentiating implicitly:

2x - 2yy' + 2xy' + 2y = 0

Tangent line will be parallel to x-axis when y' = 0.

y' = (2x + 2y)/(2y - 2x) = 0

Tangent line will be parallel to x-axis at points on the curve where x = -y:

y^2 - y^2 - 2y^2 + 4 = 0

y = +/- √2

So the points where the tangent is parallel to the x-axis are (√2, -√2) and (-√2, √2).

- RaffaeleLv 78 years ago
parallel to x-axis means slope is 0

slope is the derivative wrt x of the equation

x² - y² + 2xy + 4 = 0 (°)

2x - 2yy' + 2y + 2xy' = 0

y' = (x + y)/(y - x)

y' = 0 if x + y = 0

y = - x

x² - x² - 2x² + 4 = 0

x² = 2

x = ± √2

(√2, - √2) & (-√2, √2)

plug into (°)

plug into (°)

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