If the line y = 2x + k is a normal to the par | vedclass

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(A) The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is $y + tx = 2at + at^3$. Here,$4a = 4$,so $a = 1$. Thus,the nor

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$k = -12, t = -2$

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(A) The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is $y + tx = 2at + at^3$. Here,$4a = 4$,so $a = 1$. Thus,the normal at $(t^2, 2t)$ is $y + tx = 2t + t^3$. Comparing this with the given line $y = -tx + (2t + t^3)$,we have the slope $-t = 2$,which implies $t = -2$. Substituting $t = -2$ into the equation $y = 2x + k$,we get $y = -2x + (2(-2) + (-2)^3) = -2x + (-4 - 8) = -2x - 12$. Wait,the given line is $y = 2x + k$. Comparing $y = -tx + (2t + t^3)$ with $y = 2x + k$,we get $-t = 2 \implies t = -2$. Then $k = 2t + t^3 = 2(-2) + (-2)^3 = -4 - 8 = -12$. Therefore,$k = -12$ and $t = -2$.

If the line $y = 2x + k$ is a normal to the parabola $y^2 = 4x$ at the point $(t^2, 2t)$,then:

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