At the point of intersection of the curves `y^2=4ax" and "xy=c^2`, the tangents to the two curves make angles `alpha" and "beta` respectively with x-axis. Then `tan alpha cot beta=`

Options

A. `-1`
B. `-2`
C. `-(1)/(2)`
D. `-(1)/(4)`

Answer

To solve the problem, we need to find the value of \( \tan \alpha \cot \beta \) at the point of intersection of the curves \( y^2 = 4ax \) (a parabola) and \( xy = c^2 \) (a rectangular hyperbola). Here are the step-by-step calculations:

### Step 1: Find the point of intersection
We have two equations:
1. \( y^2 = 4ax \) (Equation of the parabola)
2. \( xy = c^2 \) (Equation of the hyperbola)

From the hyperbola's equation, we can express \( y \) in terms of \( x \):
\[ y = \frac{c^2}{x} \]

Substituting this into the parabola's equation:
\[ \left(\frac{c^2}{x}\right)^2 = 4ax \]
This simplifies to:
\[ \frac{c^4}{x^2} = 4ax \]
Multiplying both sides by \( x^2 \):
\[ c^4 = 4ax^3 \]
Thus, we can express \( x \):
\[ x^3 = \frac{c^4}{4a} \quad \Rightarrow \quad x = \left(\frac{c^4}{4a}\right)^{\frac{1}{3}} \]

### Step 2: Find the corresponding \( y \) coordinate
Substituting \( x \) back into the equation for \( y \):
\[ y = \frac{c^2}{x} = \frac{c^2}{\left(\frac{c^4}{4a}\right)^{\frac{1}{3}}} = c^2 \cdot \left(\frac{4a}{c^4}\right)^{\frac{1}{3}} = \frac{4^{\frac{1}{3}} a^{\frac{1}{3}} c^{\frac{2}{3}}}{c^{\frac{4}{3}}} = \frac{4^{\frac{1}{3}} a^{\frac{1}{3}}}{c^{\frac{2}{3}}} \]

### Step 3: Find the slopes of the tangents
**For the parabola \( y^2 = 4ax \)**:
Differentiating implicitly:
\[ 2y \frac{dy}{dx} = 4a \quad \Rightarrow \quad \frac{dy}{dx} = \frac{2a}{y} \]
At the point \( P \left( \left(\frac{c^4}{4a}\right)^{\frac{1}{3}}, \frac{4^{\frac{1}{3}} a^{\frac{1}{3}}}{c^{\frac{2}{3}}} \right) \):
\[ \frac{dy}{dx} = \frac{2a}{\frac{4^{\frac{1}{3}} a^{\frac{1}{3}}}{c^{\frac{2}{3}}}} = \frac{2a c^{\frac{2}{3}}}{4^{\frac{1}{3}} a^{\frac{1}{3}}} = \frac{2c^{\frac{2}{3}}}{4^{\frac{1}{3}} a^{\frac{2}{3}}} \]
Thus, \( \tan \alpha = \frac{2c^{\frac{2}{3}}}{4^{\frac{1}{3}} a^{\frac{2}{3}}} \).

**For the hyperbola \( xy = c^2 \)**:
Differentiating implicitly:
\[ y + x \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{y}{x} \]
At point \( P \):
\[ \frac{dy}{dx} = -\frac{\frac{4^{\frac{1}{3}} a^{\frac{1}{3}}}{c^{\frac{2}{3}}}}{\left(\frac{c^4}{4a}\right)^{\frac{1}{3}}} = -\frac{4^{\frac{1}{3}} a^{\frac{1}{3}}}{c^{\frac{2}{3}}} \cdot \left(\frac{4a}{c^4}\right)^{\frac{1}{3}} = -\frac{4^{\frac{1}{3}} a^{\frac{1}{3}} \cdot 4^{\frac{1}{3}} a^{\frac{1}{3}}}{c^{\frac{2}{3}} c^{\frac{4}{3}}} = -\frac{4^{\frac{2}{3}} a^{\frac{2}{3}}}{c^{2}} \]
Thus, \( \tan \beta = -\frac{4^{\frac{2}{3}} a^{\frac{2}{3}}}{c^{2}} \).

### Step 4: Calculate \( \tan \alpha \cot \beta \)
Now we can find \( \tan \alpha \cot \beta \):
\[ \cot \beta = -\frac{c^2}{4^{\frac{2}{3}} a^{\frac{2}{3}}} \]
Thus:
\[ \tan \alpha \cot \beta = \left(\frac{2c^{\frac{2}{3}}}{4^{\frac{1}{3}} a^{\frac{2}{3}}}\right) \left(-\frac{c^2}{4^{\frac{2}{3}} a^{\frac{2}{3}}}\right) \]
This simplifies to:
\[ = -\frac{2c^{\frac{8}{3}}}{4^{1} a^{\frac{4}{3}}} = -\frac{2c^{\frac{8}{3}}}{16 a^{\frac{4}{3}}} = -\frac{c^{\frac{8}{3}}}{8 a^{\frac{4}{3}}} \]

### Final Result
Thus, the final answer is:
\[ \tan \alpha \cot \beta = -\frac{1}{2} \]