If a normal chord of a parabola y^2 = 4ax sub | vedclass

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(D) The equation of the normal to the parabola $y^2 = 4ax$ at point $P(at^2, 2at)$ is $y + tx = 2at + at^3$.Let this normal meet the parabola again at

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$\pm \sqrt{2}$

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(D) The equation of the normal to the parabola $y^2 = 4ax$ at point $P(at^2, 2at)$ is $y + tx = 2at + at^3$.Let this normal meet the parabola again at point $Q$. The origin $O(0,0)$ is the vertex of the parabola.The combined equation of the lines $OP$ and $OQ$ is obtained by homogenizing the parabola equation $y^2 = 4ax$ using the normal equation $\frac{y + tx}{2at + at^3} = 1$:$y^2 = 4ax \left( \frac{y + tx}{2at + at^3} ight)$$y^2(2at + at^3) = 4ax(y + tx)$$4atx^2 + 4axy - (2at + at^3)y^2 = 0$Since $OP$ and $OQ$ are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$4at - (2at + at^3) = 0$$2at - at^3 = 0$$at(2 - t^2) = 0$Since $t 
eq 0$ for a normal chord,we have $t^2 = 2$,so $t = \pm \sqrt{2}$.The equation of the normal is $y = -tx + 2at + at^3$. The slope of this normal is $m = -t$.Therefore,the slope $m = \mp \sqrt{2}$,which is equivalent to $\pm \sqrt{2}$.

If a normal chord of a parabola $y^2 = 4ax$ subtends a right angle at the origin,then the slope of that normal chord is

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