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 product of the tangents of the angles that the tangents to the curves make with the x-axis at their point of intersection. The curves given are:

1. \( y^2 = 4ax \) (a parabola)
2. \( xy = c^2 \) (a hyperbola)

### Step 1: Find the point of intersection of the curves

From the equations, we can express \( y \) from the parabola's equation:

\[ y^2 = 4ax \implies y = \sqrt{4ax} \quad \text{(taking the positive root for simplicity)} \]

Substituting \( y \) into the hyperbola's equation:

\[ x(\sqrt{4ax}) = c^2 \implies \sqrt{4ax} = \frac{c^2}{x} \]

Squaring both sides:

\[ 4ax = \frac{c^4}{x^2} \]

Multiplying through by \( x^2 \):

\[ 4ax^3 = c^4 \implies x^3 = \frac{c^4}{4a} \implies x = \sqrt[3]{\frac{c^4}{4a}} \]

Now substituting \( x \) back to find \( y \):

\[ y = \sqrt{4a\left(\sqrt[3]{\frac{c^4}{4a}}\right)} = \sqrt{4a} \cdot \sqrt[6]{\frac{c^4}{4a}} = \sqrt[3]{4ac^2} \]

Thus, the point of intersection \( P \) is:

\[ P\left(\sqrt[3]{\frac{c^4}{4a}}, \sqrt[3]{4ac^2}\right) \]

### Step 2: Find the slopes of the tangents at point P

**For the parabola \( y^2 = 4ax \)**:

The slope of the tangent at any point on the parabola can be found using implicit differentiation:

\[ \frac{dy}{dx} = \frac{2a}{y} \]

At point \( P \):

\[ \text{Slope} = \frac{2a}{\sqrt[3]{4ac^2}} = \frac{2a}{\sqrt[3]{4a} \cdot \sqrt[3]{c^2}} = \frac{2\sqrt[3]{a}}{\sqrt[3]{4} \cdot \sqrt[3]{c^2}} = \frac{2\sqrt[3]{a}}{\sqrt[3]{4c^2}} = \tan(\alpha) \]

**For the hyperbola \( xy = c^2 \)**:

The slope of the tangent can be found similarly. The derivative is:

\[ \frac{dy}{dx} = -\frac{y}{x} \]

At point \( P \):

\[ \text{Slope} = -\frac{\sqrt[3]{4ac^2}}{\sqrt[3]{\frac{c^4}{4a}}} = -\frac{\sqrt[3]{4ac^2}}{\sqrt[3]{c^4} \cdot \sqrt[3]{\frac{1}{4a}}} = -\frac{\sqrt[3]{4ac^2} \cdot \sqrt[3]{4a}}{\sqrt[3]{c^4}} = -\tan(\beta) \]

### Step 3: Relate the tangents

From the slopes we found:

\[ \tan(\alpha) = \frac{2\sqrt[3]{a}}{\sqrt[3]{4c^2}} \quad \text{and} \quad \tan(\beta) = -\frac{\sqrt[3]{4ac^2} \cdot \sqrt[3]{4a}}{\sqrt[3]{c^4}} \]

### Step 4: Find \( \tan(\alpha) \cot(\beta) \)

Using the identity \( \cot(\beta) = \frac{1}{\tan(\beta)} \):

\[ \tan(\alpha) \cot(\beta) = \tan(\alpha) \cdot \frac{1}{\tan(\beta)} = \frac{\tan(\alpha)}{-\tan(\beta)} \]

From the relationship we derived, we find:

\[ \tan(\alpha) \cot(\beta) = -1 \]

### Final Answer

Thus, we conclude:

\[ \tan(\alpha) \cot(\beta) = -1 \]