Answer:
y = x + 0.5
Step-by-step explanation:
This is a very trivial exercise, follow the steps below:
Step 1: Perform the implicit differentiation of the given equation
[tex]x^2 + y^2 = (4x^2 + 2y^2 - x)^2[/tex]
[tex]2x + 2y \frac{dy}{dx} = 2(4x^2 + 2y^2 - x) ( 8x + 4y\frac{dy}{dx} - 1)\\\\[/tex]
Step 2: Make dy/dx the subject of the formula, this will be the slope of the curve:
[tex]x + y \frac{dy}{dx} = (4x^2 + 2y^2 - x) ( 8x + 4y\frac{dy}{dx} - 1)\\\\x + y \frac{dy}{dx} = 32x^3 + 16x^2y \frac{dy}{dx} - 4x^2 + 16xy^2 + 8y^3\frac{dy}{dx} - 2y^2 - 8x^2 - 4xy\frac{dy}{dx} + x \\\\\frac{dy}{dx}(y + 4xy - 8y^3) = 32x^3 - 12x^2 + 16xy^2 - 2y^2\\\\\frac{dy}{dx} = \frac{32x^3 - 12x^2 + 16xy^2 - 2y^2}{y + 4xy - 8y^3}[/tex]
Step 3: Find dy/dx at the point (0, 0.5)
[tex]\frac{dy}{dx}|(0,0.5) = \frac{32(0)^3 - 12(0)^2 + 16(0)(0.5)^2 - 2(0.5)^2}{(0.5) + 4(0)(0.5) - 8(0.5)^3}\\\\\frac{dy}{dx}|(0,0.5) =\frac{-0.5}{-0.5} \\\\\frac{dy}{dx}|(0,0.5) =1\\\\m = \frac{dy}{dx}|(0,0.5) =1[/tex]
Step 4: The equation of the tangent line to a curve at a given point is given by the equation:
[tex]y - y_1 = m(x-x_1)\\\\y - 0.5 = 1(x - 0)\\\\y = x + 0.5[/tex]
