The Cartesian equation of the parabola x = -2 | vedclass

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Question

(B) Given parametric equations are $x = -2 + 2t^2$ and $y = 2 + 4t$. From $y = 2 + 4t$,we get $t = \frac{y - 2}{4}$. Substituting this value of $t$ in

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given parametric equations are $x = -2 + 2t^2$ and $y = 2 + 4t$. From $y = 2 + 4t$,we get $t = \frac{y - 2}{4}$. Substituting this value of $t$ into the equation $x = -2 + 2t^2$: $x = -2 + 2 \left( \frac{y - 2}{4} \right)^2$ $x = -2 + 2 \left( \frac{(y - 2)^2}{16} \right)$ $x = -2 + \frac{(y - 2)^2}{8}$ Multiplying by $8$: $8x = -16 + (y - 2)^2$ $8x = -16 + y^2 - 4y + 4$ $8x = y^2 - 4y - 12$ Rearranging the terms: $y^2 - 8x - 4y - 12 = 0$. Thus,option $B$ is correct.

The Cartesian equation of the parabola $x = -2 + 2t^2$,$y = 2 + 4t$ is

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