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

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$36$

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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 $(at_1^2, 2at_1), (at_2^2, 2at_2), (at_3^2, 2at_3)$ 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,$\left(\frac{AM \cdot BN}{CD}ight)^2 = 6^2 = 36$.

List-$I$	List-$II$
$A$. Equation of the tangent drawn at $(2, \sqrt{8})$ on the curve $y^2 = 4x$ is	$(i) -36$
$B$. Equation of the normal to the curve $y^2 = 16x$,that makes an angle of $45^{\circ}$ with its axis is	$(ii) 4$
$C$. The chord joining the points $(x_1, y_1)$ and $(x_2, y_2)$ on the curve $y^2 = 12x$ is a focal chord if $y_1 y_2 =$	$(iii) 8$
$D$. $A$ value of $k$ for which $x - 3 = 0$ is the directrix of the curve $y^2 - kx + 16 = 0$ is	$(iv) x - \sqrt{2}y + 2 = 0$
	$(v) x + y - 12 = 0$
	$(vi) x - y - 12 = 0$

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