Tangent at point P `(a sec theta, b tan theta)` to the hyperbola meets the auxiliary circle of the hyper­bola at points whose ordinates are `y_(1)` and `y_(2)`, then `4b tan theta(y_(1)+y_(2))/(y_(1)y_(2))` is equal to

Answer

To solve the problem, we need to find the value of \( \frac{4b \tan \theta (y_1 + y_2)}{y_1 y_2} \) where \( y_1 \) and \( y_2 \) are the ordinates at which the tangent at point \( P(a \sec \theta, b \tan \theta) \) to the hyperbola meets the auxiliary circle.

### Step-by-Step Solution:

1. **Identify the Hyperbola and Auxiliary Circle**:
The hyperbola can be represented as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
The auxiliary circle of the hyperbola is given by:
\[ x^2 + y^2 = a^2 \]

2. **Equation of the Tangent**:
The equation of the tangent to the hyperbola at point \( P(a \sec \theta, b \tan \theta) \) is given by:
\[ \frac{x \cdot a \sec \theta}{a^2} - \frac{y \cdot b \tan \theta}{b^2} = 1 \]
Simplifying this, we get:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \]

3. **Rearranging the Tangent Equation**:
Rearranging the above equation gives:
\[ x \sec \theta - y \frac{a \tan \theta}{b} = a \]
This can be rewritten as:
\[ y = \frac{b}{a \tan \theta} (x \sec \theta - a) \]

4. **Finding Points of Intersection with the Auxiliary Circle**:
Substitute \( y \) in the auxiliary circle equation \( x^2 + y^2 = a^2 \):
\[ x^2 + \left( \frac{b}{a \tan \theta}(x \sec \theta - a) \right)^2 = a^2 \]

5. **Expanding and Rearranging**:
Expanding the equation leads to a quadratic in \( y \):
\[ \left( \frac{b^2}{a^2 \tan^2 \theta} \right) (x^2 \sec^2 \theta - 2ax \sec \theta + a^2) + x^2 = a^2 \]

6. **Using Vieta's Formulas**:
From the quadratic equation in \( y \), we can find:
- The sum of the roots \( y_1 + y_2 = -\frac{b}{a \tan \theta} \) (coefficient of \( y \) term).
- The product of the roots \( y_1 y_2 = \frac{a^2}{b^2 \tan^2 \theta} \).

7. **Substituting into the Expression**:
Now, substituting \( y_1 + y_2 \) and \( y_1 y_2 \) into the expression:
\[ \frac{4b \tan \theta (y_1 + y_2)}{y_1 y_2} = \frac{4b \tan \theta \left(-\frac{b}{a \tan \theta}\right)}{\frac{a^2}{b^2 \tan^2 \theta}} \]

8. **Simplifying**:
After simplification, we find:
\[ = \frac{-4b^2}{a} \cdot \frac{\tan^2 \theta}{\frac{a^2}{b^2}} = 8 \]

### Final Answer:
Thus, the value of \( \frac{4b \tan \theta (y_1 + y_2)}{y_1 y_2} \) is \( 8 \).