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(D) Given the parabola equation: $x^2-4x-4y+16=0$. Differentiating with respect to $x$: $2x-4-4\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = \frac{2x-4}

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-$5$

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(D) Given the parabola equation: $x^2-4x-4y+16=0$. Differentiating with respect to $x$: $2x-4-4\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = \frac{2x-4}{4} = \frac{x-2}{2}$. The slope of the tangent $2x-y-5=0$ is $m=2$. Equating the derivative to the slope: $\frac{x-2}{2} = 2$ $\Rightarrow x-2=4$ $\Rightarrow x=6$. Substituting $x=6$ into the tangent equation: $2(6)-y-5=0$ $\Rightarrow 12-y-5=0$ $\Rightarrow y=7$. Thus,the point of contact is $P(6, 7)$. The slope of the normal at $P$ is $m' = -\frac{1}{m} = -\frac{1}{2}$. The equation of the normal is $(y-7) = -\frac{1}{2}(x-6)$. $2y-14 = -x+6 \Rightarrow x+2y-20=0$. Dividing by $2$ to match the form $ax+y+c=0$: $\frac{1}{2}x+y-10=0$. Comparing with $ax+y+c=0$,we get $a=\frac{1}{2}$ and $c=-10$. Therefore,$ac = \frac{1}{2} 	imes (-10) = -5$.

Let the equation of the tangent at a point $P$ on the parabola $x^2-4x-4y+16=0$ be $2x-y-5=0$. If the equation of the normal drawn at $P$ to this parabola is $ax+y+c=0$,then find the value of $ac$.

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