If the curves y^2=6x and 9x^2+by^2=16 interse | vedclass

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(D) For the curve $C_1: y^2=6x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 6$,so $\left(\frac{dy}{dx}\right)_{C_1} = \frac{3}{y}$.F

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$\frac{9}{2}$

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(D) For the curve $C_1: y^2=6x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 6$,so $\left(\frac{dy}{dx}ight)_{C_1} = \frac{3}{y}$.For the curve $C_2: 9x^2+by^2=16$,differentiating with respect to $x$ gives $18x + 2by \frac{dy}{dx} = 0$,so $\left(\frac{dy}{dx}ight)_{C_2} = -\frac{9x}{by}$.Since the curves intersect at right angles,the product of their slopes must be $-1$:$\left(\frac{3}{y}ight) 	imes \left(-\frac{9x}{by}ight) = -1$$\Rightarrow \frac{27x}{by^2} = 1$$\Rightarrow 27x = by^2$Substituting $y^2=6x$ into the equation:$27x = b(6x)$Since $x 
eq 0$ at the point of intersection,we divide by $3x$:$9 = 2b$$b = \frac{9}{2}$

$(a)$ Placenta	$(i)$ Androgens
$(b)$ Zona pellucida	$(ii)$ Human Chorionic Gonadotropin $(hCG)$
$(c)$ Bulbourethral glands	$(iii)$ Layer of the ovum
$(d)$ Leydig cells	$(iv)$ Lubrication of the penis

If the curves $y^2=6x$ and $9x^2+by^2=16$ intersect each other at right angles,the value of $b$ is

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