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

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(C) Given line: $x-2y+k=0 \Rightarrow y = \frac{1}{2}x + \frac{k}{2}$.Comparing with $y=mx+c$,we get $m=\frac{1}{2}$ and $c=\frac{k}{2}$.Given parabol

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$7$

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(C) Given line: $x-2y+k=0 \Rightarrow y = \frac{1}{2}x + \frac{k}{2}$.Comparing with $y=mx+c$,we get $m=\frac{1}{2}$ and $c=\frac{k}{2}$.Given parabola: $y^2-4x-4y+8=0$.Rewriting in standard form: $(y-2)^2 = 4x-8+4 = 4(x-1)$.Let $Y = y-2$ and $X = x-1$,then $Y^2 = 4X$.The condition for the line $Y=mX+c'$ to be a tangent to $Y^2=4aX$ is $c' = \frac{a}{m}$.Here $a=1$,$m=\frac{1}{2}$,and $Y = \frac{1}{2}(X+1) - 2 = \frac{1}{2}X - \frac{3}{2}$.So $c' = -\frac{3}{2}$.Setting $c' = \frac{a}{m} = \frac{1}{1/2} = 2$ is not applicable directly due to the shift.Alternatively,substitute $x = 2y-k$ into the parabola equation:$(y-2)^2 = 4(2y-k-1) \Rightarrow y^2-4y+4 = 8y-4k-4$.$y^2-12y+(8+4k) = 0$.For tangency,the discriminant $D=0$:$(-12)^2 - 4(1)(8+4k) = 0 \Rightarrow 144 - 32 - 16k = 0$.$112 = 16k \Rightarrow k = 7$.

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$,then the value of $k$ is

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