If the normal at (ap^2, 2ap) on the parabola | vedclass

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Question

(A) The equation of the normal to the parabola $y^2 = 4ax$ at point $(ap^2, 2ap)$ is $y = -px + 2ap + ap^3$.Since this normal meets the parabola again

Vedclass · Accepted Answer

$p^2 + pq + 2 = 0$

Vedclass · Answer

(A) The equation of the normal to the parabola $y^2 = 4ax$ at point $(ap^2, 2ap)$ is $y = -px + 2ap + ap^3$.Since this normal meets the parabola again at $(aq^2, 2aq)$,the point $(aq^2, 2aq)$ must satisfy the normal equation.Substituting $x = aq^2$ and $y = 2aq$ into the equation:$2aq = -p(aq^2) + 2ap + ap^3$Dividing by $a$ (assuming $a 
eq 0$):$2q = -pq^2 + 2p + p^3$$pq^2 - 2q + p^3 - 2p = 0$$pq^2 - 2q + p(p^2 - 2) = 0$Since $p 
eq q$,we can simplify the relation between $p$ and $q$ as $q = -p - \frac{2}{p}$.Multiplying by $p$,we get $pq = -p^2 - 2$,which rearranges to $p^2 + pq + 2 = 0$.

If the normal at $(ap^2, 2ap)$ on the parabola $y^2 = 4ax$ meets the parabola again at $(aq^2, 2aq)$,then:

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