Key Concepts and Formulas

• Orthogonal Intersection: Two curves intersect orthogonally if the product of the slopes of their tangents at the point of intersection is -1. That is, $m_1m_2 = -1$.
• Implicit Differentiation: Used to find the derivative of a function defined implicitly (e.g., $f(x, y) = 0$).
• Chain Rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$.

Step-by-Step Solution

Step 1: Define the Point of Intersection

Let $(x_1, y_1)$ be the point of intersection of the curves $y^2 = 6x$ and $9x^2 + by^2 = 16$. This point must satisfy both equations. Therefore, we have:

$y_1^2 = 6x_1 \quad (1)$ $9x_1^2 + by_1^2 = 16 \quad (2)$

Step 2: Find the Slope of the Tangent to the First Curve

The first curve is $y^2 = 6x$. We use implicit differentiation to find $\frac{dy}{dx}$:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(6x)$ $2y \frac{dy}{dx} = 6$ $\frac{dy}{dx} = \frac{3}{y}$

The slope of the tangent at $(x_1, y_1)$ is: $m_1 = \frac{3}{y_1}$

Step 3: Find the Slope of the Tangent to the Second Curve

The second curve is $9x^2 + by^2 = 16$. We use implicit differentiation to find $\frac{dy}{dx}$:

$\frac{d}{dx}(9x^2 + by^2) = \frac{d}{dx}(16)$ $18x + 2by \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{18x}{2by} = -\frac{9x}{by}$

The slope of the tangent at $(x_1, y_1)$ is: $m_2 = -\frac{9x_1}{by_1}$

Step 4: Apply the Orthogonality Condition

Since the curves intersect orthogonally, $m_1 m_2 = -1$:

$\left(\frac{3}{y_1}\right)\left(-\frac{9x_1}{by_1}\right) = -1$ $-\frac{27x_1}{by_1^2} = -1$ $\frac{27x_1}{by_1^2} = 1$ $27x_1 = by_1^2 \quad (3)$

Step 5: Solve for b

We have $y_1^2 = 6x_1$ from equation (1). Substitute this into equation (3):

$27x_1 = b(6x_1)$

Since $(x_1, y_1)$ is a point of intersection, $x_1$ cannot be zero. If $x_1 = 0$, then $y_1 = 0$, and substituting into the second equation, we get $0 = 16$, which is a contradiction. Therefore, $x_1 \neq 0$. We can divide both sides by $x_1$:

$27 = 6b$ $b = \frac{27}{6} = \frac{9}{2}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs when differentiating and applying the orthogonality condition.
• Forgetting Chain Rule: When differentiating $y^2$ with respect to $x$, remember to use the chain rule: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.
• Assuming x1 = 0: Remember to check if $x_1$ can be zero before dividing by it.

Summary

We found the slopes of the tangents to the two curves using implicit differentiation. Then, using the orthogonality condition, we derived an equation relating $x_1$, $y_1$, and $b$. Finally, we used the fact that $(x_1, y_1)$ satisfies the equation of the first curve to eliminate $x_1$ and $y_1$ and solve for $b$. The value of $b$ is $\frac{9}{2}$.

Final Answer

The final answer is \boxed{9/2}, which corresponds to option (A).