The equation of the tangent at (-4, -4) on th | vedclass

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Question

(D) Given the curve equation $x^2 = -4y$. Differentiating both sides with respect to $x$,we get: $2x = -4 \frac{dy}{dx}$ $\frac{dy}{dx} = -\frac{x}{2}

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$2x - y + 4 = 0$

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(D) Given the curve equation $x^2 = -4y$. Differentiating both sides with respect to $x$,we get: $2x = -4 \frac{dy}{dx}$ $\frac{dy}{dx} = -\frac{x}{2}$ Now,find the slope of the tangent at the point $(-4, -4)$: $m = \left( \frac{dy}{dx} \right)_{(-4, -4)} = -\frac{-4}{2} = 2$. The equation of the tangent line at $(x_1, y_1)$ is given by $(y - y_1) = m(x - x_1)$. Substituting the values $m = 2$,$x_1 = -4$,and $y_1 = -4$: $y - (-4) = 2(x - (-4))$ $y + 4 = 2(x + 4)$ $y + 4 = 2x + 8$ $2x - y + 4 = 0$.

The equation of the tangent at $(-4, -4)$ on the curve $x^2 = -4y$ is

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