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

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Question

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

Vedclass · Accepted Answer

$2x - y + 4 = 0$

Vedclass · Answer

(B) Given the curve equation is $x^{2} = -4y$. Differentiating both sides with respect to $x$,we get $2x = -4 \frac{dy}{dx}$. Thus,$\frac{dy}{dx} = -\frac{x}{2}$. The slope of the tangent at point $P(-4, -4)$ is $m = -\frac{-4}{2} = 2$. The equation of the tangent line passing through $P(x_1, y_1) = (-4, -4)$ with slope $m = 2$ is given by $y - y_1 = m(x - x_1)$. Substituting the values,we get $y - (-4) = 2(x - (-4))$. $y + 4 = 2(x + 4)$. $y + 4 = 2x + 8$. $2x - y + 4 = 0$. Therefore,the correct option is $B$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

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

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