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

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(D) Let the point of intersection be $(x_1, y_1)$.For the curve $y^2 = 6x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 6$,so $\frac{

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$\frac{9}{2}$

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(D) Let the point of intersection be $(x_1, y_1)$.For the curve $y^2 = 6x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 6$,so $\frac{dy}{dx} = \frac{3}{y_1}$. Let this be $m_1 = \frac{3}{y_1}$.For the curve $9x^2 + by^2 = 16$,differentiating with respect to $x$ gives $18x + 2by \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{9x_1}{by_1}$. Let this be $m_2 = -\frac{9x_1}{by_1}$.Since the curves intersect at right angles,$m_1 	imes m_2 = -1$,which implies $(\frac{3}{y_1}) 	imes (-\frac{9x_1}{by_1}) = -1$,so $\frac{27x_1}{by_1^2} = 1$. Since $y_1^2 = 6x_1$,we have $\frac{27x_1}{b(6x_1)} = 1$,which simplifies to $\frac{27}{6b} = 1$,so $b = \frac{27}{6} = \frac{9}{2}$.

Column-$I$	Column-$II$
$A$. Circle	$P$. Locus of point $(h, k)$ such that the line $hx + ky = 1$ touches the circle $x^2 + y^2 = 4$
$B$. Parabola	$Q$. Point $z$ in the complex plane satisfies $\|z + 2\| - \|z - 2\| = \pm 3$
$C$. Hyperbola	$R$. Eccentricity of the conic lies in the interval $[1, \infty)$
	$S$. Point $z$ in the complex plane satisfies $Re(z + 1)^2 = \|z\|^2 + 1$

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

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