The parametric equations of the parabola y^2 | vedclass

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(A) Given equation: $y^2 - 2y - 12x - 11 = 0$Complete the square for $y$: $(y^2 - 2y + 1) - 1 - 12x - 11 = 0$$(y - 1)^2 = 12x + 12$$(y - 1)^2 = 12(x +

Vedclass · Accepted Answer

$x = 3t^2 - 1, y = 6t + 1$

Vedclass · Answer

(A) Given equation: $y^2 - 2y - 12x - 11 = 0$Complete the square for $y$: $(y^2 - 2y + 1) - 1 - 12x - 11 = 0$$(y - 1)^2 = 12x + 12$$(y - 1)^2 = 12(x + 1)$Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get $h = -1, k = 1$,and $4a = 12 \implies a = 3$.The parametric equations for $(y - k)^2 = 4a(x - h)$ are $x = h + at^2$ and $y = k + 2at$.Substituting the values: $x = -1 + 3t^2$ and $y = 1 + 2(3)t = 1 + 6t$.Thus,$x = 3t^2 - 1$ and $y = 6t + 1$.

The parametric equations of the parabola $y^2 - 12x - 2y - 11 = 0$ are:

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