If the length of sub-normal is equal to the length of sub-tangent at any point (3,4) on the curve `y=f(x)` and the tangent at (3,4) to `y=f(x)` meets the coordinate axes at `Aa n dB` , then the maximum area of the triangle `O A B ,` where `O` is origin, is 45/2 (b) 49/2 (c) 25/2 (d) 81/2

Options

A. 45/2
B. 49/2
C. 25/2
D. 81/2

Answer

To solve the problem step by step, we will follow the information provided in the video transcript and derive the necessary equations and areas.

### Step 1: Understanding the Problem
We need to find the maximum area of triangle OAB formed by the tangent to the curve \( y = f(x) \) at the point (3, 4), where the lengths of the sub-normal and sub-tangent are equal.

### Step 2: Sub-tangent and Sub-normal
The lengths of the sub-tangent \( ST \) and sub-normal \( SN \) at a point on the curve are given by:
- Sub-tangent \( ST = \frac{y}{f'(x)} \)
- Sub-normal \( SN = \frac{y f'(x)}{1} \)

Given that \( ST = SN \), we have:
\[ \frac{y}{f'(x)} = y f'(x) \]
This simplifies to:
\[ 1 = (f'(x))^2 \]
Thus, \( f'(x) = \pm 1 \).

### Step 3: Finding the Derivative at Point (3, 4)
At the point (3, 4), we can take \( f'(3) = 1 \) or \( f'(3) = -1 \).

### Step 4: Equation of the Tangent Line
Using the point-slope form of the equation of a line, the equation of the tangent at (3, 4) is:
1. For \( f'(3) = 1 \):
\[ y - 4 = 1(x - 3) \implies y = x + 1 \]
2. For \( f'(3) = -1 \):
\[ y - 4 = -1(x - 3) \implies y = -x + 7 \]

### Step 5: Finding Points A and B
1. For \( y = x + 1 \):
- The x-intercept (A) occurs when \( y = 0 \):
\[ 0 = x + 1 \implies x = -1 \quad \text{(Point A: (-1, 0))} \]
- The y-intercept (B) occurs when \( x = 0 \):
\[ y = 0 + 1 = 1 \quad \text{(Point B: (0, 1))} \]

2. For \( y = -x + 7 \):
- The x-intercept (A) occurs when \( y = 0 \):
\[ 0 = -x + 7 \implies x = 7 \quad \text{(Point A: (7, 0))} \]
- The y-intercept (B) occurs when \( x = 0 \):
\[ y = -0 + 7 = 7 \quad \text{(Point B: (0, 7))} \]

### Step 6: Area of Triangle OAB
The area \( A \) of triangle OAB can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

1. For the first tangent \( y = x + 1 \):
\[ \text{Area} = \frac{1}{2} \times (-1) \times 1 = \frac{1}{2} \text{ square units} \]

2. For the second tangent \( y = -x + 7 \):
\[ \text{Area} = \frac{1}{2} \times 7 \times 7 = \frac{49}{2} \text{ square units} \]

### Conclusion
The maximum area of triangle OAB is:
\[ \frac{49}{2} \]

### Final Answer
The correct option is (b) \( \frac{49}{2} \).