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Q.
If a chord which is normal to the parabola at one end and subtends a right angle at the vertex, then slope of the chord is
Conic Sections
Solution:
Chord $PQ$ is normal to the parabola at Qand is equation is $y = mx -2 am - am ^{3}$
The slope of normal $PQ$ is $m =\tan \theta .$ The joint equation of $OP$ and $OQ$ is obtained by making the equation of parabola $y^{2}=4$ ax homogeneous with the help of $(1)$ Thus joint equation of lines $OP , OQ$ is
$y ^{2}=4 ax \left\{\frac{ m x - y }{2 am + am ^{3}}\right\}$
or $m \left(2+ m ^{2}\right) y ^{2}+4 xy -4 mx ^{2}=0$
Since $\angle POQ =\frac{\pi}{2}$, the sum of the coefficients of $x ^{2}$ and $y ^{2}$ in $(2)$ must be zero.
$m\left(2+m^{2}\right)-4 m=0$ or $m\left(m^{2}-2\right)=0$
As $m \neq 0$, we get $m^{2}=2$ or $m=\pm \sqrt{2}$. Thus, $m=\sqrt{2}$.
