(b) Find the slope of the following curves at the given points. (i) \( y^{2}+x^{2}=y^{4}-2 x \) at \( (-2,1) \) and \( (-2,-1) \)

To find the slope of the curve defined by the equation \( y^2 + x^2 = y^4 - 2x \) at the given points, we will use implicit differentiation. 1. **Differentiate both sides of the equation with respect to \( x \)**: The equation is: \[ y^2 + x^2 = y^4 - 2x \] Differentiating both sides: - The left side: \[ \frac{d}{dx}(y^2) + \frac{d}{dx}(x^2) = 2y \frac{dy}{dx} + 2x \] - The right side: \[ \frac{d}{dx}(y^4) - \frac{d}{dx}(2x) = 4y^3 \frac{dy}{dx} - 2 \] Setting the derivatives equal gives: \[ 2y \frac{dy}{dx} + 2x = 4y^3 \frac{dy}{dx} - 2 \] 2. **Rearranging the equation**: \[ 2y \frac{dy}{dx} - 4y^3 \frac{dy}{dx} = -2 - 2x \] \[ (2y - 4y^3) \frac{dy}{dx} = -2 - 2x \] \[ \frac{dy}{dx} = \frac{-2 - 2x}{2y - 4y^3} \] \[ \frac{dy}{dx} = \frac{-1 - x}{y - 2y^3} \] 3. **Evaluate the derivative at the points \((-2, 1)\) and \((-2, -1)\)**: - **At the point \((-2, 1)\)**: \[ \frac{dy}{dx} = \frac{-1 - (-2)}{1 - 2(1)^3} = \frac{-1 + 2}{1 - 2} = \frac{1}{-1} = -1 \] - **At the point \((-2, -1)\)**: \[ \frac{dy}{dx} = \frac{-1 - (-2)}{-1 - 2(-1)^3} = \frac{-1 + 2}{-1 + 2} = \frac{1}{1} = 1 \] 4. **Final results**: - The slope of the curve at the point \((-2, 1)\) is \(-1\). - The slope of the curve at the point \((-2, -1)\) is \(1\).