The tangent at any point on the curve `x = at^3. y = at^4` divides the abscissa of the point of contact in the ratio m:n, then `|n + m|` is equal to (m and n are co-prime)

Options

A. `1//4`
B. `3//4`
C. `3//2`
D. `2//5`

Explanation

To solve the problem, we need to find the value of \(|n + m|\) where the tangent at any point on the curve \(x = at^3\) and \(y = at^4\) divides the abscissa of the point of contact in the ratio \(m:n\).

### Step-by-Step Solution:

1. **Find the derivatives**:
The parametric equations are given as:
\[ x = at^3, \quad y = at^4 \]
We need to find \(\frac{dy}{dx}\):
\[ \frac{dx}{dt} = 3at^2, \quad \frac{dy}{dt} = 4at^3 \]
Therefore,
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4at^3}{3at^2} = \frac{4t}{3} \]

2. **Equation of the tangent line**:
The equation of the tangent line at the point \((at^3, at^4)\) is given by:
\[ y - at^4 = \frac{4t}{3}(x - at^3) \]
Rearranging this gives:
\[ y = \frac{4t}{3}x - \frac{4t}{3}at^3 + at^4 \]

3. **Finding the x-intercept**:
To find the x-intercept, set \(y = 0\):
\[ 0 = \frac{4t}{3}x - \frac{4t}{3}at^3 + at^4 \]
Rearranging gives:
\[ \frac{4t}{3}x = \frac{4t}{3}at^3 - at^4 \]
Factoring out \(\frac{4t}{3}\):
\[ x = \frac{at^3}{4} \]

4. **Identifying the points**:
Let \(O\) be the origin \((0, 0)\), \(A\) be the point of contact \((at^3, at^4)\), and \(B\) be the x-intercept \(\left(\frac{at^3}{4}, 0\right)\).

5. **Finding the ratio \(m:n\)**:
The lengths \(OA\) and \(AB\) can be calculated as follows:
- Length \(OA = at^3\)
- Length \(AB = at^3 - \frac{at^3}{4} = \frac{3at^3}{4}\)

The ratio \(m:n\) is given by:
\[ \frac{OA}{AB} = \frac{at^3}{\frac{3at^3}{4}} = \frac{4}{3} \]
Thus, \(m = 4\) and \(n = 3\).

6. **Finding \(|n + m|\)**:
Now, we compute:
\[ |n + m| = |3 + 4| = 7 \]

### Final Answer:
\[ |n + m| = 7 \]