If -xy^3 - 1 - y = 0,

Question

then find the equation of all tangent lines to curve when y=1.

Answer 1

If -xy³ – 1 – y = 0, then find the equation of all tangent lines to curve when y = 1.

Detailed Solution – Using Implicit Differentiation

Note: when y = 1 ==> -x(y)³ – 1 – (y) = 0 ==> -x(1)³ – 1 – (1) = 0 ==> -x – 1 – 1 = 0 ==> x = -2

Therefore, (x, y) = (-2, 1)

-xy³ – 1 – y = 0 ==> d/dx(-xy³ – 1 – y = 0) ==> d/dx(-xy³) – d/dx(1) – d/dx(y) = d/dx(0) ==>

(-x)(3y²)dy/dx + (y³)(-1) – 0 – dy/dx = 0==>(-3xy²)dy/dx – (y³) – dy/dx = 0==>-(3xy² + 1)dy/dx = y³==>

dy/dx = -y³/(3xy² + 1)

Slope = m = dy/dx_{(-2, 1)} = -(y)³/(3(x)(y)² + 1) = -(1)³/(3(-2)(1)² + 1) = -1/(-6 + 1) = -1/(-5) ==> m = 1/5

Equation of tangent at (x, y) = (-2, 1): y = mx + b

y = mx + b ==> y = (1/5)x + b ==> 1 = (1/5)(-2) + b ==> b = 1 + (2/5) ==> b = 7/5

Therefore, Equation of tangent at (x, y) = (-2, 1): y = mx + b ==> y = (1/5)x + 7/5

Answer 2

There is just one tangent line to the graph at (-2,1)

y = (1/5)x + 7/5