If a normal chord at a point t on the parabol | vedclass

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(C) 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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$2$

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(C) 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 combined equation of the lines $OP$ and $OQ$ joining the vertex $O(0,0)$ to the points $P$ and $Q$ is obtained by homogenizing the equation of the parabola $y^2=4ax$ using the equation of the normal: $y^2 = 4ax \left( \frac{y+tx}{2at+at^3} ight)$ $y^2(2at + at^3) = 4ax(y + tx)$ $y^2(2at + at^3) = 4axy + 4atx^2$ $4atx^2 + 4axy - (2at + at^3)y^2 = 0$ Since $OP$ and $OQ$ are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $4at - (2at + at^3) = 0$ $4at - 2at - at^3 = 0$ $2at - at^3 = 0$ $at(2 - t^2) = 0$ Since $t 
eq 0$ for a normal chord,we have $t^2 = 2$.

If a normal chord at a point $t$ on the parabola $y^2=4ax$ subtends a right angle at the vertex,then $t^2$ equals to

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