If the tangent at any point P on the curve `x^m y^n = a^(m+n), mn != 0` meets the coordinate axes in A, B then show that `AP:BP` is a constant.

Options

A. `m:n`
B. `n:m`
C. `-m:n`
D. `-n:m`

Answer

To solve the problem, we need to show that the ratio \( AP:BP \) is a constant for the tangent line to the curve \( x^m y^n = a^{m+n} \) at any point \( P(h, k) \).

### Step 1: Find the equation of the tangent line at point \( P(h, k) \)

1. **Differentiate the curve**:
We start with the equation of the curve:
\[ x^m y^n = a^{m+n} \]
Differentiating both sides with respect to \( x \) using implicit differentiation gives:
\[ m x^{m-1} y^n + n x^m y^{n-1} \frac{dy}{dx} = 0 \]
Rearranging for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{m}{n} \cdot \frac{y}{x} \]

2. **Evaluate the slope at point \( P(h, k) \)**:
Substituting \( x = h \) and \( y = k \):
\[ \frac{dy}{dx} \bigg|_{(h, k)} = -\frac{m}{n} \cdot \frac{k}{h} \]

3. **Write the equation of the tangent line**:
Using the point-slope form of the line:
\[ y - k = -\frac{m}{n} \cdot \frac{k}{h}(x - h) \]
Rearranging gives:
\[ y = -\frac{m}{n} \cdot \frac{k}{h} x + \left(k + \frac{m}{n} \cdot \frac{k}{h} h\right) \]
Simplifying:
\[ y = -\frac{m}{n} \cdot \frac{k}{h} x + k \left(1 + \frac{m}{n}\right) \]

### Step 2: Find the points where the tangent meets the axes

1. **Find point A (where the tangent meets the y-axis)**:
Set \( x = 0 \):
\[ y = k \left(1 + \frac{m}{n}\right) \]
Thus, point \( A \) is:
\[ A(0, k(1 + \frac{m}{n})) \]

2. **Find point B (where the tangent meets the x-axis)**:
Set \( y = 0 \):
\[ 0 = -\frac{m}{n} \cdot \frac{k}{h} x + k \left(1 + \frac{m}{n}\right) \]
Solving for \( x \):
\[ x = \frac{n}{m} h (1 + \frac{m}{n}) = h \frac{n + m}{m} \]
Thus, point \( B \) is:
\[ B\left(h \frac{n + m}{m}, 0\right) \]

### Step 3: Calculate the distances \( AP \) and \( BP \)

1. **Distance \( AP \)**:
\[ AP = \sqrt{(h - 0)^2 + \left(k - k(1 + \frac{m}{n})\right)^2} \]
Simplifying:
\[ AP = \sqrt{h^2 + k^2 \left(-\frac{m}{n}\right)^2} = \sqrt{h^2 + \frac{m^2 k^2}{n^2}} \]

2. **Distance \( BP \)**:
\[ BP = \sqrt{\left(h - h \frac{n + m}{m}\right)^2 + (k - 0)^2} \]
Simplifying:
\[ BP = \sqrt{\left(h(1 - \frac{n + m}{m})\right)^2 + k^2} = \sqrt{\left(h \frac{m - n}{m}\right)^2 + k^2} \]

### Step 4: Find the ratio \( \frac{AP}{BP} \)

Now we can find the ratio:
\[ \frac{AP}{BP} = \frac{\sqrt{h^2 + \frac{m^2 k^2}{n^2}}}{\sqrt{\left(h \frac{m - n}{m}\right)^2 + k^2}} \]

### Conclusion

The ratio \( \frac{AP}{BP} \) simplifies to a constant value that does not depend on the specific point \( P(h, k) \) on the curve, thus proving that \( AP:BP \) is a constant.