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

Options

A. 2
B. 4
C. `9//2`
D. None

Answer

To find the value of \( b \) for which the curves \( y^2 = 16x \) and \( 9x^2 + by^2 = 16 \) cut each other at right angles, we will follow these steps:

### Step 1: Find the slopes of the tangents to the curves at the point of intersection.
The first curve is given by:
\[ y^2 = 16x \]
Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{16}{2y} = \frac{8}{y} \]
Thus, the slope \( m_1 \) of the tangent to the first curve at the point \( (H, K) \) is:
\[ m_1 = \frac{8}{K} \]

### Step 2: Differentiate the second curve.
The second curve is given by:
\[ 9x^2 + by^2 = 16 \]
Differentiating both sides with respect to \( x \):
\[ 18x + 2by \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{18x}{2by} = -\frac{9x}{by} \]
Thus, the slope \( m_2 \) of the tangent to the second curve at the point \( (H, K) \) is:
\[ m_2 = -\frac{9H}{bK} \]

### Step 3: Use the condition for the curves to intersect at right angles.
For the curves to intersect at right angles, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \frac{8}{K} \cdot \left(-\frac{9H}{bK}\right) = -1 \]
This simplifies to:
\[ -\frac{72H}{bK^2} = -1 \implies \frac{72H}{bK^2} = 1 \implies 72H = bK^2 \]

### Step 4: Express \( H \) in terms of \( K \) using the first curve.
From the first curve \( y^2 = 16x \), substituting \( K \) for \( y \) and \( H \) for \( x \):
\[ K^2 = 16H \implies H = \frac{K^2}{16} \]

### Step 5: Substitute \( H \) into the equation from Step 3.
Substituting \( H = \frac{K^2}{16} \) into \( 72H = bK^2 \):
\[ 72 \left(\frac{K^2}{16}\right) = bK^2 \]
This simplifies to:
\[ \frac{72K^2}{16} = bK^2 \implies \frac{72}{16} = b \implies b = \frac{72}{16} = \frac{9}{2} \]

### Conclusion
Thus, the value of \( b \) is:
\[ \boxed{\frac{9}{2}} \]