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

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(D) Given equation: $y^2-4x-8y-12=0$ Rearranging terms: $y^2-8y = 4x+12$ Completing the square: $y^2-8y+16 = 4x+12+16$ $(y-4)^2 = 4x+28$ $(y-4)^2 = 4(

Vedclass · Accepted Answer

$x=-7+t^2, y=4+2t$

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(D) Given equation: $y^2-4x-8y-12=0$ Rearranging terms: $y^2-8y = 4x+12$ Completing the square: $y^2-8y+16 = 4x+12+16$ $(y-4)^2 = 4x+28$ $(y-4)^2 = 4(x+7)$ Let $Y = y-4$ and $X = x+7$. Then the equation becomes $Y^2 = 4X$. Comparing with the standard form $Y^2 = 4aX$,we get $4a = 4$,so $a = 1$. The parametric equations for $Y^2 = 4aX$ are $X = at^2$ and $Y = 2at$. Substituting $a = 1$: $X = t^2$ and $Y = 2t$. Substituting back $X = x+7$ and $Y = y-4$: $x+7 = t^2 \Rightarrow x = -7+t^2$ $y-4 = 2t \Rightarrow y = 4+2t$ Thus,the correct option is $D$.

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

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