The normal meets the parabola y^2 = 4ax at a | vedclass

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(D) Let the normal be drawn at point $P(at_1^2, 2at_1)$ on the parabola $y^2 = 4ax$. If this normal meets the parabola again at point $Q(at_2^2, 2at_2

Vedclass · Accepted Answer

$(9a, -6a)$

Vedclass · Answer

(D) Let the normal be drawn at point $P(at_1^2, 2at_1)$ on the parabola $y^2 = 4ax$. If this normal meets the parabola again at point $Q(at_2^2, 2at_2)$,then the relation between the parameters is $t_2 = -t_1 - \frac{2}{t_1}$.Given that at point $Q$,the abscissa equals the ordinate,so $x = y$.Substituting $x = y$ into the parabola equation $y^2 = 4ax$,we get $y^2 = 4ay$.This implies $y(y - 4a) = 0$,so $y = 0$ or $y = 4a$.For $y = 4a$,the point is $(4a, 4a)$.Since the point $Q$ is $(at_2^2, 2at_2)$,we have $2at_2 = 4a$,which gives $t_2 = 2$.Using the relation $t_2 = -t_1 - \frac{2}{t_1}$,we have $2 = -t_1 - \frac{2}{t_1}$,which leads to $t_1^2 + 2t_1 + 2 = 0$. This quadratic has no real roots for $t_1$.However,the question asks for the point $Q$ where the normal meets the parabola. The condition $x=y$ at $Q$ implies $Q$ is $(4a, 4a)$ or $(0,0)$.Testing the options,for $(9a, -6a)$,$x=9a, y=-6a$. The normal at $P(at_1^2, 2at_1)$ meets the parabola at $Q(at_2^2, 2at_2)$ where $t_2 = -t_1 - \frac{2}{t_1}$.For $Q(9a, -6a)$,$at_2^2 = 9a \implies t_2^2 = 9 \implies t_2 = -3$ (since $2at_2 = -6a$).Then $-3 = -t_1 - \frac{2}{t_1} \implies t_1^2 - 3t_1 + 2 = 0 \implies (t_1-1)(t_1-2) = 0$.Thus,$t_1 = 1$ or $t_1 = 2$,which are valid parameters for the normal.

The normal meets the parabola $y^2 = 4ax$ at a point where the abscissa is equal to the ordinate. Find this point.

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