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

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(C) Let the curves intersect each other at point $P(x_1, y_1)$.Since the point of intersection lies on both curves,we have:$y_1^2 = 6x_1 \quad \dots(i

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$\frac{9}{2}$

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(C) Let the curves intersect each other at point $P(x_1, y_1)$.Since the point of intersection lies on both curves,we have:$y_1^2 = 6x_1 \quad \dots(i)$$9x_1^2 + by_1^2 = 16 \quad \dots(ii)$Now,find the slope of the tangent to both curves at the point of intersection $P(x_1, y_1)$.For curve $(i)$,differentiating with respect to $x$:$2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$So,$m_1 = \left( \frac{dy}{dx} \right)_{P} = \frac{3}{y_1}$.For curve $(ii)$,differentiating with respect to $x$:$18x + 2by \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{9x}{by}$So,$m_2 = \left( \frac{dy}{dx} \right)_{P} = -\frac{9x_1}{by_1}$.Since the curves intersect at right angles,$m_1 m_2 = -1$:$\left( \frac{3}{y_1} \right) \left( -\frac{9x_1}{by_1} \right) = -1$$\frac{27x_1}{by_1^2} = 1 \Rightarrow b = \frac{27x_1}{y_1^2}$Substitute $y_1^2 = 6x_1$ from equation $(i)$ into the expression for $b$:$b = \frac{27x_1}{6x_1} = \frac{27}{6} = \frac{9}{2}$.

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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