For the parabola y=2+4t, x=-2+2t^2,the ends o | vedclass

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(C) Given parametric equations are $y=2+4t$ and $x=-2+2t^2$. From the first equation,$t = \frac{y-2}{4}$. Substituting this into the second equation:

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$-1$

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(C) Given parametric equations are $y=2+4t$ and $x=-2+2t^2$. From the first equation,$t = \frac{y-2}{4}$. Substituting this into the second equation: $x = -2 + 2\left(\frac{y-2}{4}\right)^2 = -2 + 2\frac{(y-2)^2}{16} = -2 + \frac{(y-2)^2}{8}$. Rearranging gives $(y-2)^2 = 8(x+2)$. Comparing this with $(y-k)^2 = 4a(x-h)$,we get $h=-2, k=2$,and $4a=8$,so $a=2$. The ends of the latus rectum are at $(h+a, k \pm 2a)$,which are $(-2+2, 2 \pm 4)$,i.e.,$(0, 6)$ and $(0, -2)$. For $y=6$,$2+4t=6 \implies 4t=4 \implies t=1$. For $y=-2$,$2+4t=-2 \implies 4t=-4 \implies t=-1$. Thus,$\alpha=1$ and $\beta=-1$. Therefore,$\alpha \beta = (1)(-1) = -1$.

For the parabola $y=2+4t, x=-2+2t^2$,the ends of the latus rectum are at $t=\alpha$ and $t=\beta$. Then $\alpha \beta=$

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