If the tangent at any point on the curve `x^4 + y^4 = c^4` cuts off intercepts a and b on the coordinate axes, the value of `a^(-4/3)+b^(-4/3)` is

Options

A. ` c^(-4//3)`
B. ` c^(-1//2)`
C. ` c^(1//2)`
D. none of these

Explanation

To solve the problem, we need to find the value of \( a^{-\frac{4}{3}} + b^{-\frac{4}{3}} \) where \( a \) and \( b \) are the x-intercept and y-intercept of the tangent to the curve defined by \( x^4 + y^4 = c^4 \).

### Step-by-Step Solution:

1. **Identify the curve and point on the curve**:
The given curve is \( x^4 + y^4 = c^4 \). Let the point on the curve be \( (x_0, y_0) \). Therefore, we have:
\[ x_0^4 + y_0^4 = c^4 \]

2. **Differentiate the equation**:
Differentiate the equation \( x^4 + y^4 = c^4 \) implicitly with respect to \( x \):
\[ \frac{d}{dx}(x^4) + \frac{d}{dx}(y^4) = 0 \]
This gives:
\[ 4x^3 + 4y^3 \frac{dy}{dx} = 0 \]
Simplifying, we find:
\[ \frac{dy}{dx} = -\frac{x^3}{y^3} \]

3. **Find the slope at the point \( (x_0, y_0) \)**:
At the point \( (x_0, y_0) \), the slope of the tangent line is:
\[ \text{slope} = -\frac{x_0^3}{y_0^3} \]

4. **Write the equation of the tangent line**:
The equation of the tangent line at the point \( (x_0, y_0) \) can be expressed as:
\[ y - y_0 = -\frac{x_0^3}{y_0^3}(x - x_0) \]
Rearranging gives:
\[ y y_0^3 - y_0^4 = -x_0^3 x + x_0^4 \]
Rearranging further, we have:
\[ x_0^3 x + y_0^3 y = x_0^4 + y_0^4 \]

5. **Find the x-intercept \( a \)**:
To find the x-intercept \( a \), set \( y = 0 \):
\[ x_0^3 a = x_0^4 + y_0^4 \]
Thus,
\[ a = \frac{x_0^4 + y_0^4}{x_0^3} = \frac{c^4}{x_0^3} \]

6. **Find the y-intercept \( b \)**:
To find the y-intercept \( b \), set \( x = 0 \):
\[ y_0^3 b = x_0^4 + y_0^4 \]
Thus,
\[ b = \frac{x_0^4 + y_0^4}{y_0^3} = \frac{c^4}{y_0^3} \]

7. **Calculate \( a^{-\frac{4}{3}} + b^{-\frac{4}{3}} \)**:
Now substituting the values of \( a \) and \( b \):
\[ a^{-\frac{4}{3}} = \left(\frac{c^4}{x_0^3}\right)^{-\frac{4}{3}} = \frac{x_0^{4}}{c^{\frac{16}{3}}} \]
\[ b^{-\frac{4}{3}} = \left(\frac{c^4}{y_0^3}\right)^{-\frac{4}{3}} = \frac{y_0^{4}}{c^{\frac{16}{3}}} \]
Therefore,
\[ a^{-\frac{4}{3}} + b^{-\frac{4}{3}} = \frac{x_0^4 + y_0^4}{c^{\frac{16}{3}}} \]

8. **Substitute \( x_0^4 + y_0^4 \)**:
Since \( x_0^4 + y_0^4 = c^4 \), we have:
\[ a^{-\frac{4}{3}} + b^{-\frac{4}{3}} = \frac{c^4}{c^{\frac{16}{3}}} = c^{4 - \frac{16}{3}} = c^{\frac{12}{3} - \frac{16}{3}} = c^{-\frac{4}{3}} \]

### Final Result:
Thus, the value of \( a^{-\frac{4}{3}} + b^{-\frac{4}{3}} \) is:
\[ \boxed{c^{-\frac{4}{3}}} \]