The parametric equations of the parabola y^2- | vedclass

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(C) Given equation of the parabola is $y^2-8x-4y-12=0$. Completing the square for $y$: $(y^2-4y+4)-4-8x-12=0$ $(y-2)^2=8x+16$ $(y-2)^2=8(x+2)$. Compar

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$x=-2+2t^2, y=2+4t$

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(C) Given equation of the parabola is $y^2-8x-4y-12=0$. Completing the square for $y$: $(y^2-4y+4)-4-8x-12=0$ $(y-2)^2=8x+16$ $(y-2)^2=8(x+2)$. Comparing this with the standard form $(y-k)^2=4a(x-h)$,we get $h=-2, k=2$,and $4a=8$,which implies $a=2$. The parametric equations for $(y-k)^2=4a(x-h)$ are $x=h+at^2$ and $y=k+2at$. Substituting the values,we get $x=-2+2t^2$ and $y=2+2(2)t$,which simplifies to $x=-2+2t^2$ and $y=2+4t$. Thus,the correct option is $C$.

The parametric equations of the parabola $y^2-8x-4y-12=0$ are

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