If m is the slope of the normal to the parabo | vedclass

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(D) The given equation of the parabola is $y^2 + 4y = -4x$. Completing the square,we get $(y + 2)^2 = -4x + 4$,which simplifies to $(y + 2)^2 = -4(x -

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$2$

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(D) The given equation of the parabola is $y^2 + 4y = -4x$. Completing the square,we get $(y + 2)^2 = -4x + 4$,which simplifies to $(y + 2)^2 = -4(x - 1)$. Comparing this with $(y - k)^2 = -4a(x - h)$,we have $h = 1$,$k = -2$,and $4a = 4$,so $a = 1$. The focus of the parabola is $(h - a, k) = (1 - 1, -2) = (0, -2)$. The point given is $P(-3, 2)$. The slope of the tangent at $P(x_1, y_1)$ for $y^2 + 4x + 4y = 0$ is found by differentiating: $2y \frac{dy}{dx} + 4 + 4 \frac{dy}{dx} = 0$. Substituting $y = 2$,we get $2(2) \frac{dy}{dx} + 4 + 4 \frac{dy}{dx} = 0$,which implies $8 \frac{dy}{dx} = -4$,so $\frac{dy}{dx} = -1/2$. The slope of the normal $m$ is the negative reciprocal of the slope of the tangent: $m = -1 / (-1/2) = 2$.

If $m$ is the slope of the normal to the parabola $y^2 + 4x + 4y = 0$ at the point $(-3, 2)$,which is one of the endpoints of a focal chord,find the value of $m$.

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