The normal at the point (2, 4) to the parabol | vedclass

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(A) For the parabola $y^2 = 4ax$,the point $(2, 4)$ lies on it,so $4^2 = 4a(2) \implies 16 = 8a \implies a = 2$. Comparing $(2, 4)$ with $(at^2, 2at)$

Vedclass · Accepted Answer

$(18, -12)$

Vedclass · Answer

(A) For the parabola $y^2 = 4ax$,the point $(2, 4)$ lies on it,so $4^2 = 4a(2) \implies 16 = 8a \implies a = 2$. Comparing $(2, 4)$ with $(at^2, 2at)$,we get $2t = 4 \implies t_1 = 2$. The normal at $t_1$ meets the parabola at $t_2 = -t_1 - \frac{2}{t_1}$. Substituting $t_1 = 2$,we get $t_2 = -2 - \frac{2}{2} = -2 - 1 = -3$. The coordinates of the point are $(at_2^2, 2at_2) = (2(-3)^2, 2(2)(-3)) = (2 	imes 9, -12) = (18, -12)$.

The normal at the point $(2, 4)$ to the parabola $y^2 = 4ax$ meets the parabola again at another point. Find the coordinates of this point.

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