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

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(A) Given equation of parabola is $x^2-8 x+12 y+15=0$. Completing the square for $x$: $x^2-8 x+16 = -12 y - 15 + 16$ $(x-4)^2 = -12 y + 1$ $(x-4)^2 =

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$x=4+6 t, y=\frac{1}{12}-3 t^2$

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(A) Given equation of parabola is $x^2-8 x+12 y+15=0$. Completing the square for $x$: $x^2-8 x+16 = -12 y - 15 + 16$ $(x-4)^2 = -12 y + 1$ $(x-4)^2 = -12(y - \frac{1}{12})$. Comparing this with the standard form $(x-h)^2 = -4a(y-k)$,we get $h=4$,$k=\frac{1}{12}$,and $4a=12$,so $a=3$. The parametric equations for $(x-h)^2 = -4a(y-k)$ are $x = h + 2at$ and $y = k - at^2$. Substituting the values,we get $x = 4 + 2(3)t = 4 + 6t$ and $y = \frac{1}{12} - 3t^2$. Thus,the correct option is $A$.

The parametric equations of the parabola $x^2-8 x+12 y+15=0$ are

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