# let x^2 + 4xy + y^2 + 3 = 0

are there any points on the curve where the tangent is horizontal or vertical? justify your answer.

## 2 x dx + 4 x dy + 4 y dx + 2 y dy = 0

dx (2x+4y) + dy(2y+4x) = 0

dy/dx = -(2x+4y)/(4x+2y)

if the top is 0, tangent horizontal

if the bottom is zero, vertical

for example for yertical

4x=-2y

y = -2x

plug that back in the original equation

x^2 +4x(-2x) + (-2x)^2 + 3 = 0

x^2 -8x^2 +4x^2 = -3

-3 x^2 = -3

x = +/- 1

Be sure to check my arithmetic and do the other half of the problem for numerator = 0