Let A, B and C be three points on the parabol | vedclass

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(B) For the parabola $y^2=4ax$,we have $4a=6$,so $a=\frac{3}{2}$. Let the points $A, B, C$ be $(at_1^2, 2at_1), (at_2^2, 2at_2), (at_3^2, 2at_3)$ resp

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) For the parabola $y^2=4ax$,we have $4a=6$,so $a=\frac{3}{2}$. Let the points $A, B, C$ be $(at_1^2, 2at_1), (at_2^2, 2at_2), (at_3^2, 2at_3)$ respectively.Line $L$ passes through $C$ and is parallel to the $x$-axis,so its equation is $y=2at_3$.The line $AB$ has the equation $y(t_1+t_2)=2x+2at_1t_2$.Point $D$ is the intersection of $AB$ and $L$,so $2at_3(t_1+t_2)=2x_D+2at_1t_2$,which gives $x_D=a(t_1t_3+t_2t_3-t_1t_2)$.$AM = |2at_1 - 2at_3| = |2a(t_1-t_3)|$.$BN = |2at_2 - 2at_3| = |2a(t_2-t_3)|$.$CD = |x_D - at_3^2| = |a(t_1t_3+t_2t_3-t_1t_2-t_3^2)| = a|(t_3-t_1)(t_3-t_2)|$.Thus,$\frac{AM \cdot BN}{CD} = \frac{|2a(t_1-t_3)| \cdot |2a(t_2-t_3)|}{a|(t_3-t_1)(t_3-t_2)|} = 4a$.Since $a=\frac{3}{2}$,we have $4a = 4 	imes \frac{3}{2} = 6$.Therefore,$\left(\frac{AM \cdot BN}{CD}ight)^2 = 6^2 = 36$.

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