The length of the normal chord of the parabol | vedclass

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(B) The equation of the normal to the parabola $y^2 = 4ax$ (where $a=1$) at parameter $t$ is $y + tx = 2at + at^3$,which simplifies to $y + tx = 2t +

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$8\sqrt{2}$

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(B) The equation of the normal to the parabola $y^2 = 4ax$ (where $a=1$) at parameter $t$ is $y + tx = 2at + at^3$,which simplifies to $y + tx = 2t + t^3$.The slope of this normal is $-t$. Given that the normal makes an angle of $\frac{\pi}{4}$ with the $x$-axis,its slope is $	an(\frac{\pi}{4}) = 1$.Therefore,$-t = 1$,which gives $t = -1$.The coordinates of point $P$ are $(at^2, 2at) = (1, -2)$.If the normal at $t_1$ meets the parabola again at $t_2$,then $t_2 = -t_1 - \frac{2}{t_1}$.Substituting $t_1 = -1$,we get $t_2 = -(-1) - \frac{2}{-1} = 1 + 2 = 3$.The coordinates of point $Q$ are $(at_2^2, 2at_2) = (3^2, 2(3)) = (9, 6)$.The length of the normal chord $PQ$ is the distance between $P(1, -2)$ and $Q(9, 6)$:$PQ = \sqrt{(9-1)^2 + (6 - (-2))^2} = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$.

The length of the normal chord of the parabola $y^2 = 4x$,which makes an angle of $\frac{\pi}{4}$ with the $x$-axis,is:

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