If the tangentat P of the curve `y^2 = x^3` intersect the curve again at Q and the straigta line `OP, OQ` have inclinations `alpha and beta` where O is origin, then `tanalpha/tan beta` has the value equals to

Options

A. `-1`
B. `-2`
C. 2
D. `sqrt2`

Answer

To solve the problem, we need to find the value of \(\frac{\tan \alpha}{\tan \beta}\) where \(\alpha\) and \(\beta\) are the inclinations of the lines \(OP\) and \(OQ\) respectively, and \(P\) and \(Q\) are points on the curve \(y^2 = x^3\).

### Step-by-Step Solution:

1. **Identify the curve and points:**
The curve given is \(y^2 = x^3\). Let's assume the point \(P\) on the curve has coordinates \((t_1^2, t_1^3)\) where \(t_1\) is a parameter.

2. **Find the slope of the tangent at point \(P\):**
To find the slope of the tangent at point \(P\), we differentiate \(y^2 = x^3\) implicitly:
\[ 2y \frac{dy}{dx} = 3x^2 \implies \frac{dy}{dx} = \frac{3x^2}{2y} \]
At point \(P\), substituting \(x = t_1^2\) and \(y = t_1^3\):
\[ \frac{dy}{dx} \bigg|_P = \frac{3(t_1^2)^2}{2(t_1^3)} = \frac{3t_1^4}{2t_1^3} = \frac{3t_1}{2} \]

3. **Equation of the tangent line at point \(P\):**
The equation of the tangent line at point \(P\) can be expressed as:
\[ y - t_1^3 = \frac{3t_1}{2}(x - t_1^2) \]
Rearranging gives:
\[ y = \frac{3t_1}{2}x - \frac{3t_1^3}{2} + t_1^3 = \frac{3t_1}{2}x - \frac{t_1^3}{2} \]

4. **Find the intersection point \(Q\):**
To find where this tangent intersects the curve again, substitute \(y\) in the curve equation:
\[ \left(\frac{3t_1}{2}x - \frac{t_1^3}{2}\right)^2 = x^3 \]
Expanding and simplifying will yield a cubic equation in \(x\). The point \(P\) corresponds to one root of this equation, and we can find the other root, which we denote as \(t_2\).

5. **Relate the slopes to angles:**
The slopes of the lines \(OP\) and \(OQ\) are given by:
\[ \tan \alpha = \frac{y_1}{x_1} = \frac{t_1^3}{t_1^2} = t_1 \]
\[ \tan \beta = \frac{y_2}{x_2} = \frac{t_2^3}{t_2^2} = t_2 \]

6. **Find the ratio \(\frac{\tan \alpha}{\tan \beta}\):**
We have:
\[ \frac{\tan \alpha}{\tan \beta} = \frac{t_1}{t_2} \]
From the relationship derived from the cubic equation, we find that:
\[ t_2 = -\frac{1}{2} t_1 \]
Thus,
\[ \frac{\tan \alpha}{\tan \beta} = \frac{t_1}{-\frac{1}{2} t_1} = -2 \]

### Final Answer:
\[ \frac{\tan \alpha}{\tan \beta} = -2 \]