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

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(B) Let the point $P$ on the parabola $y^2=4ax$ be $(at^2, 2at)$. The equation of the normal at $P$ is $y + tx = 2at + at^3$.To find the combined equa

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$2$

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(B) Let the point $P$ on the parabola $y^2=4ax$ be $(at^2, 2at)$. The equation of the normal at $P$ is $y + tx = 2at + at^3$.To find the combined equation of the lines joining the vertex $O(0,0)$ to the points of intersection of the normal and the parabola,we homogenize the equation of the parabola using the equation of the normal.From the normal equation,we have $1 = \frac{y + tx}{2at + at^3}$.Substituting this into the parabola equation $y^2 = 4ax(1)$:$y^2 = 4ax \left( \frac{y + tx}{2at + at^3} ight)$$y^2(2at + at^3) = 4axy + 4atx^2$$4atx^2 + 4axy - (2at + at^3)y^2 = 0$Since the lines $OP$ 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$Since $a 
eq 0$ and $t 
eq 0$ (for a chord to exist),we divide by $at$:$2 - t^2 = 0 \Rightarrow t^2 = 2$.

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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