Q97 / 160 · TS EAMCET 2021 (04 Aug Shift 1) · Mathematics

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

Solution

$y^{2} - 4 x - 8 y - 12 = 0$

$y^{2} - 8 y = 4 x + 12$

$y^{2} - 8 y + 16 - 16 = 4 x + 12$

${(y - 4)}^{2} = 4 x + 28$

${(y - 4)}^{2} = 4 (x + 7)$

Let $y - 4 = Y, x + 7 = X$

$\Rightarrow Y^{2} = 4 X$

On comparing it with $y^{2} = 4 a x$ , we get

$4 a = 4$

$a = 1$

Parametric equations are

$X = a t^{2}, Y = 2 a t$

$x + 7 = t^{2} & y - 4 = 2 t$

Hence, the correct answer is option: 'D'.