nd are inclined at avgicsTangents are drawn from the point `(alpha, beta)` to the hyperbola `3x^2- 2y^2=6` and are inclined atv angle `theta and phi` to the x-axis.If `tan theta.tan phi=2`, prove that `beta^2 = 2alpha^2 - 7`.

Explanation

To solve the problem, we need to prove that \( \beta^2 = 2\alpha^2 - 7 \) given the conditions of the hyperbola and the tangents drawn from the point \((\alpha, \beta)\).

### Step 1: Write the equation of the hyperbola
The hyperbola is given as:
\[ 3x^2 - 2y^2 = 6 \]
We can rewrite this in standard form:
\[ \frac{x^2}{2} - \frac{y^2}{3} = 1 \]
From this, we identify \( a^2 = 2 \) and \( b^2 = 3 \).

### Step 2: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola at the point \((\alpha, \beta)\) can be expressed as:
\[ y = mx \pm \sqrt{a^2m^2 - b^2} \]
Substituting \( a^2 \) and \( b^2 \):
\[ y = mx \pm \sqrt{2m^2 - 3} \]

### Step 3: Substitute the point \((\alpha, \beta)\) into the tangent equation
Since the point \((\alpha, \beta)\) lies on the tangent, we substitute \( x = \alpha \) and \( y = \beta \):
\[ \beta = m\alpha \pm \sqrt{2m^2 - 3 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \beta - m\alpha = \pm \sqrt{2m^2 - 3} \]
Squaring both sides:
\[ (\beta - m\alpha)^2 = 2m^2 - 3 \]
Expanding the left side:
\[ \beta^2 - 2m\alpha\beta + m^2\alpha^2 = 2m^2 - 3 \]

### Step 5: Rearranging to form a quadratic in \(m\)
Rearranging the equation gives:
\[ m^2(\alpha^2 - 2) - 2m\alpha\beta + (\beta^2 + 3) = 0 \]
This is a quadratic equation in \(m\).

### Step 6: Using the product of the slopes
Let the slopes of the tangents be \( m_1 \) and \( m_2 \). The product of the slopes is given by:
\[ m_1 m_2 = \frac{c}{a} = \frac{\beta^2 + 3}{\alpha^2 - 2} \]
Given that \( \tan \theta \tan \phi = 2 \), we have:
\[ m_1 m_2 = 2 \]
Setting these equal gives:
\[ \frac{\beta^2 + 3}{\alpha^2 - 2} = 2 \]

### Step 7: Cross-multiplying and simplifying
Cross-multiplying gives:
\[ \beta^2 + 3 = 2(\alpha^2 - 2) \]
Expanding the right side:
\[ \beta^2 + 3 = 2\alpha^2 - 4 \]
Rearranging gives:
\[ \beta^2 = 2\alpha^2 - 7 \]

### Conclusion
We have proved that:
\[ \beta^2 = 2\alpha^2 - 7 \]