If x-2y+k=0 is a tangent to the parabola y^2- | vedclass

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(D) The given parabola is $y^2-4y-4x+8=0$. Completing the square,we get $(y-2)^2 = 4x-4$,which simplifies to $(y-2)^2 = 4(x-1)$. The equation of a tan

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$\frac{2}{5}$

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(D) The given parabola is $y^2-4y-4x+8=0$. Completing the square,we get $(y-2)^2 = 4x-4$,which simplifies to $(y-2)^2 = 4(x-1)$. The equation of a tangent to the parabola $(y-k_0)^2 = 4a(x-h_0)$ with slope $m$ is $(y-k_0) = m(x-h_0) + \frac{a}{m}$. Here,$h_0=1, k_0=2, a=1$. So,$y-2 = m(x-1) + \frac{1}{m}$,which gives $y = mx - m + \frac{1}{m} + 2$. The given tangent is $x-2y+k=0$,which can be written as $y = \frac{1}{2}x + \frac{k}{2}$. Comparing the slopes,$m = \frac{1}{2}$. Comparing the intercepts,$\frac{k}{2} = -m + \frac{1}{m} + 2 = -\frac{1}{2} + 2 + 2 = \frac{7}{2}$,so $k=7$. The point on the parabola is $(1, 7)$. Differentiating the parabola $y^2-4x-4y+8=0$ with respect to $x$: $2y \frac{dy}{dx} - 4 - 4 \frac{dy}{dx} = 0$. $\frac{dy}{dx}(2y-4) = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y-2}$. At $(1, 7)$,the slope is $\frac{2}{7-2} = \frac{2}{5}$.

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$,then the slope of the tangent drawn at $(1, k)$ on the given parabola is

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