Step 1: Understanding the Concept:
Two curves intersect at right angles (orthogonally) if their tangents at the point of intersection are perpendicular.
This means the product of their slopes at the point of intersection must be $-1$.
Step 2: Key Formula or Approach:
Let $(x_1, y_1)$ be the point of intersection.
Find the slope $m_1 = \left.\frac{dy}{dx}\right|_{(x_1,y_1)}$ for the first curve.
Find the slope $m_2 = \left.\frac{dy}{dx}\right|_{(x_1,y_1)}$ for the second curve.
Apply the condition $m_1 \cdot m_2 = -1$.
Step 3: Detailed Explanation:
Let the curves intersect at $(x_1, y_1)$.
Curve 1: $y^2 = 6x$
Differentiating with respect to $x$:
\[ 2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y} \]
Slope at $(x_1, y_1)$ is $m_1 = \frac{3}{y_1}$.
Curve 2: $9x^2 + by^2 = 16$
Differentiating with respect to $x$:
\[ 18x + 2by \frac{dy}{dx} = 0 \Rightarrow 2by \frac{dy}{dx} = -18x \Rightarrow \frac{dy}{dx} = -\frac{9x}{by} \]
Slope at $(x_1, y_1)$ is $m_2 = -\frac{9x_1}{by_1}$.
Since they intersect orthogonally, $m_1 \cdot m_2 = -1$:
\[ \left( \frac{3}{y_1} \right) \left( -\frac{9x_1}{by_1} \right) = -1 \]
\[ -\frac{27x_1}{by_1^2} = -1 \]
\[ by_1^2 = 27x_1 \]
We also know that the intersection point $(x_1, y_1)$ lies on both curves.
From Curve 1, we have $y_1^2 = 6x_1$.
Substitute $y_1^2 = 6x_1$ into the condition $by_1^2 = 27x_1$:
\[ b(6x_1) = 27x_1 \]
Assuming the point of intersection is not the origin (if $x_1=0$, then $y_1=0$, but $(0,0)$ does not satisfy the second curve $9(0)^2+b(0)^2=16 \Rightarrow 0=16$, which is false), we can divide both sides by $x_1 \neq 0$:
\[ 6b = 27 \]
\[ b = \frac{27}{6} = \frac{9}{2} \]
Step 4: Final Answer:
The value of $b$ is $\frac{9}{2}$.