If the tangent at `(alpha,beta)` to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` . Cuts the auxiliary circle at points whose ordinates are `y_1" and "y_2`, then `(1)/(y_1)+(1)/(y_2)` is equal to :

Options

A. `(1)/(alpha)`
B. `(2)/(alpha)`
C. `(1)/(beta)`
D. `(2)/(beta)`

Answer

To solve the problem step by step, we will follow the given information about the hyperbola and the auxiliary circle.

### Step 1: Write the equation of the tangent to the hyperbola
The equation of the hyperbola is given by:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

The equation of the tangent to this hyperbola at the point \((\alpha, \beta)\) is given by:

\[ \frac{\alpha x}{a^2} - \frac{\beta y}{b^2} = 1 \]

### Step 2: Write the equation of the auxiliary circle
The auxiliary circle of the hyperbola is given by the equation:

\[ x^2 + y^2 = a^2 \]

### Step 3: Substitute the tangent equation into the auxiliary circle equation
We will express \(x\) from the tangent equation and substitute it into the auxiliary circle equation. Rearranging the tangent equation gives:

\[ \alpha x = a^2 + \frac{\beta y}{b^2} \implies x = \frac{a^2}{\alpha} + \frac{\beta y}{\alpha b^2} \]

Now substitute this expression for \(x\) into the auxiliary circle equation:

\[ \left(\frac{a^2}{\alpha} + \frac{\beta y}{\alpha b^2}\right)^2 + y^2 = a^2 \]

### Step 4: Expand and simplify the equation
Expanding the left-hand side:

\[ \left(\frac{a^2}{\alpha}\right)^2 + 2\left(\frac{a^2}{\alpha}\right)\left(\frac{\beta y}{\alpha b^2}\right) + \left(\frac{\beta y}{\alpha b^2}\right)^2 + y^2 = a^2 \]

This leads to:

\[ \frac{a^4}{\alpha^2} + 2\frac{a^2\beta y}{\alpha^2 b^2} + \frac{\beta^2 y^2}{\alpha^2 b^4} + y^2 = a^2 \]

### Step 5: Rearranging the equation
Rearranging gives us a quadratic equation in terms of \(y\):

\[ \left(1 + \frac{\beta^2}{\alpha^2 b^4}\right)y^2 + 2\frac{a^2\beta}{\alpha^2 b^2}y + \left(\frac{a^4}{\alpha^2} - a^2\right) = 0 \]

### Step 6: Identify coefficients for sum and product of roots
Let \(y_1\) and \(y_2\) be the roots of this quadratic equation. From the quadratic formula, we know:

- The sum of the roots \(y_1 + y_2 = -\frac{b}{a}\) where \(b\) is the coefficient of \(y\) and \(a\) is the coefficient of \(y^2\).
- The product of the roots \(y_1 y_2 = \frac{c}{a}\) where \(c\) is the constant term.

Thus, we have:

\[ y_1 + y_2 = -\frac{2\frac{a^2\beta}{\alpha^2 b^2}}{1 + \frac{\beta^2}{\alpha^2 b^4}} \]

\[ y_1 y_2 = \frac{\frac{a^4}{\alpha^2} - a^2}{1 + \frac{\beta^2}{\alpha^2 b^4}} \]

### Step 7: Calculate \( \frac{1}{y_1} + \frac{1}{y_2} \)
Using the formula:

\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{y_1 + y_2}{y_1 y_2} \]

Substituting the values we found:

\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{-\frac{2\frac{a^2\beta}{\alpha^2 b^2}}{1 + \frac{\beta^2}{\alpha^2 b^4}}}{\frac{\frac{a^4}{\alpha^2} - a^2}{1 + \frac{\beta^2}{\alpha^2 b^4}}} \]

### Step 8: Simplifying the expression
This simplifies to:

\[ \frac{-2\beta}{\frac{a^4 - \alpha^2 a^2}{\alpha^2 b^2}} = \frac{-2\beta \cdot \alpha^2 b^2}{a^4 - \alpha^2 a^2} \]

### Final Result
Thus, we conclude that:

\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{2}{\beta} \]