If `y = m x +5 ` is a tangent to the curve `x ^(3) y ^(3) = ax ^(3) +by^(3)at P (1,2),` then

Options

A. (a) `a+b=(18)/(5)`
B. (b) `a gt b`
C. (c) `a lt b`
D. (d) `a +b=(19)/(5)`

Answer

To solve the problem, we need to determine the values of \( a \) and \( b \) such that the line \( y = mx + 5 \) is a tangent to the curve defined by the equation \( x^3 y^3 = ax^3 + by^3 \) at the point \( P(1, 2) \).

### Step 1: Find the slope of the curve at point \( P(1, 2) \)

The given curve is:
\[ x^3 y^3 = ax^3 + by^3 \]

We differentiate both sides with respect to \( x \) using implicit differentiation. Applying the product rule:
\[ \frac{d}{dx}(x^3 y^3) = 3x^2 y^3 + x^3 \cdot 3y^2 \frac{dy}{dx} \]
On the right side:
\[ \frac{d}{dx}(ax^3 + by^3) = 3ax^2 + 3by^2 \frac{dy}{dx} \]

Setting both derivatives equal gives:
\[ 3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx} = 3ax^2 + 3by^2 \frac{dy}{dx} \]

### Step 2: Rearranging to isolate \( \frac{dy}{dx} \)

Rearranging the equation:
\[ 3x^3 y^2 \frac{dy}{dx} - 3by^2 \frac{dy}{dx} = 3ax^2 - 3x^2 y^3 \]
Factoring out \( \frac{dy}{dx} \):
\[ \left(3x^3 y^2 - 3by^2\right) \frac{dy}{dx} = 3ax^2 - 3x^2 y^3 \]
Thus,
\[ \frac{dy}{dx} = \frac{3ax^2 - 3x^2 y^3}{3x^3 y^2 - 3by^2} \]

### Step 3: Substitute \( P(1, 2) \) into the derivative

Substituting \( x = 1 \) and \( y = 2 \):
\[ \frac{dy}{dx} = \frac{3a(1^2) - 3(1^2)(2^3)}{3(1^3)(2^2) - 3b(2^2)} = \frac{3a - 24}{12 - 12b} \]

### Step 4: Find the slope of the tangent line

The slope of the tangent line given by \( y = mx + 5 \) is \( m \). Therefore, we have:
\[ \frac{3a - 24}{12 - 12b} = m \]

### Step 5: Substitute point \( P(1, 2) \) into the curve equation

Substituting \( P(1, 2) \) into the original curve equation:
\[ 1^3 \cdot 2^3 = a(1^3) + b(2^3) \]
This simplifies to:
\[ 8 = a + 8b \quad \text{(1)} \]

### Step 6: Substitute point \( P(1, 2) \) into the tangent line equation

Substituting \( P(1, 2) \) into the tangent line equation:
\[ 2 = m(1) + 5 \]
This gives:
\[ m = 2 - 5 = -3 \quad \text{(2)} \]

### Step 7: Substitute \( m = -3 \) into the slope equation

Substituting \( m = -3 \) into the slope equation:
\[ \frac{3a - 24}{12 - 12b} = -3 \]
Cross-multiplying gives:
\[ 3a - 24 = -3(12 - 12b) \]
This simplifies to:
\[ 3a - 24 = -36 + 36b \]
Rearranging gives:
\[ 3a - 36b = -12 \quad \text{(3)} \]

### Step 8: Solve equations (1) and (3)

From equation (1):
\[ a + 8b = 8 \quad \Rightarrow \quad a = 8 - 8b \]

Substituting \( a \) into equation (3):
\[ 3(8 - 8b) - 36b = -12 \]
This simplifies to:
\[ 24 - 24b - 36b = -12 \]
Combining like terms:
\[ 24 - 60b = -12 \]
Solving for \( b \):
\[ -60b = -12 - 24 \quad \Rightarrow \quad -60b = -36 \quad \Rightarrow \quad b = \frac{36}{60} = \frac{3}{5} \]

### Step 9: Find \( a \)

Substituting \( b = \frac{3}{5} \) back into equation (1):
\[ a + 8\left(\frac{3}{5}\right) = 8 \]
This gives:
\[ a + \frac{24}{5} = 8 \quad \Rightarrow \quad a = 8 - \frac{24}{5} = \frac{40}{5} - \frac{24}{5} = \frac{16}{5} \]

### Step 10: Find \( a + b \)

Finally, we find \( a + b \):
\[ a + b = \frac{16}{5} + \frac{3}{5} = \frac{19}{5} \]

### Conclusion

Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{16}{5}, \quad b = \frac{3}{5} \]
And the final answer is:
\[ \boxed{\frac{19}{5}} \]