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 $2x = -4 \frac{dy}{dx}$.Thus,$\frac{dy}{dx} = -\frac{x}

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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 $2x = -4 \frac{dy}{dx}$.Thus,$\frac{dy}{dx} = -\frac{x}{2}$.At the point $(-4, -4)$,the slope of the tangent is $m = \left(\frac{dy}{dx}\right)_{(-4, -4)} = -\frac{-4}{2} = 2$.The equation of the tangent at $(x_1, y_1)$ is given by $(y - y_1) = m(x - x_1)$.Substituting the values,we get $(y - (-4)) = 2(x - (-4))$.$y + 4 = 2(x + 4)$.$y + 4 = 2x + 8$.$2x - y + 4 = 0$.

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

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