S equiv y^2 - 4ax = 0 and S' equiv y^2 + ax = | vedclass

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(D) Given parabolas are $S \equiv y^2 - 4ax = 0$ and $S' \equiv y^2 + 4ax = 0$ (assuming standard form $y^2 = -4ax$).Let $P$ be a point on $S' = 0$ as

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$\frac{t}{2}$

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(D) Given parabolas are $S \equiv y^2 - 4ax = 0$ and $S' \equiv y^2 + 4ax = 0$ (assuming standard form $y^2 = -4ax$).Let $P$ be a point on $S' = 0$ as $P = \left(-\frac{t^2}{4a}, t\right)$.The feet of the perpendiculars from $P$ to the axes are $A = \left(-\frac{t^2}{4a}, 0\right)$ and $B = (0, t)$.The equation of line $AB$ is $\frac{x}{-t^2/4a} + \frac{y}{t} = 1$,which simplifies to $4ax - ty + t^2 = 0$.The tangent to $S \equiv y^2 = 4ax$ at point $Q(at_1^2, 2at_1)$ is $y t_1 = x + at_1^2$,or $x - t_1 y + at_1^2 = 0$.Comparing $4ax - ty + t^2 = 0$ with $x - t_1 y + at_1^2 = 0$ (after dividing by $4a$),we get $\frac{t}{4a} = t_1$ and $\frac{t^2}{4a} = at_1^2$.Substituting $t_1 = \frac{t}{4a}$,we find the relation leads to $t_1 = \frac{t}{2}$.

$S \equiv y^2 - 4ax = 0$ and $S' \equiv y^2 + ax = 0$ are two parabolas,and $P(t)$ is a point on the parabola $S' = 0$. If $A$ and $B$ are the feet of the perpendiculars from $P$ onto the coordinate axes and $AB$ is a tangent to the parabola $S = 0$ at the point $Q(t_1)$,then $t_1 =$

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