If the tangent at point P(h, k) on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` cuts the circle `x^(2)+y^(2)=a^(2)` at points `Q(x_(1),y_(1))` and `R(x_(2),y_(2))`, then the value of `(1)/(y_(1))+(1)/(y_(2))` is

Options

A. `(1)/(k)`
B. `(2)/(k)`
C. `(ab)/(k)`
D. `(a+b)/(k)`

Explanation

To solve the problem step by step, we will derive the required expression for \( \frac{1}{y_1} + \frac{1}{y_2} \) given the tangent to the hyperbola and its intersection with the circle.

### Step 1: Equation of the Tangent to the Hyperbola
The equation of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point \( P(h, k) \) is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Substituting \( x_1 = h \) and \( y_1 = k \), we get:
\[ \frac{xh}{a^2} - \frac{yk}{b^2} = 1 \]

### Step 2: Rearranging the Tangent Equation
From the tangent equation, we can express \( x \) in terms of \( y \):
\[ \frac{xh}{a^2} = 1 + \frac{yk}{b^2} \]
Multiplying both sides by \( a^2 \):
\[ x = \frac{a^2}{h} + \frac{a^2ky}{b^2} \]

### Step 3: Substitute into the Circle's Equation
The circle's equation is \( x^2 + y^2 = a^2 \). Substituting the expression for \( x \):
\[ \left(\frac{a^2}{h} + \frac{a^2ky}{b^2}\right)^2 + y^2 = a^2 \]
Expanding the left side:
\[ \left(\frac{a^4}{h^2} + 2\frac{a^4ky}{hb^2} + \frac{a^4k^2y^2}{b^4}\right) + y^2 = a^2 \]

### Step 4: Forming a Quadratic in \( y \)
Rearranging gives us a quadratic equation in \( y \):
\[ \left(1 + \frac{a^4k^2}{b^4}\right)y^2 + 2\frac{a^4k}{hb^2}y + \left(\frac{a^4}{h^2} - a^2\right) = 0 \]

### Step 5: Identify Coefficients
Let:
- \( A = 1 + \frac{a^4k^2}{b^4} \)
- \( B = 2\frac{a^4k}{hb^2} \)
- \( C = \frac{a^4}{h^2} - a^2 \)

### Step 6: Use Vieta's Formulas
From Vieta's formulas, we know:
- \( y_1 + y_2 = -\frac{B}{A} = -\frac{2\frac{a^4k}{hb^2}}{1 + \frac{a^4k^2}{b^4}} \)
- \( y_1y_2 = \frac{C}{A} = \frac{\frac{a^4}{h^2} - a^2}{1 + \frac{a^4k^2}{b^4}} \)

### Step 7: Calculate \( \frac{1}{y_1} + \frac{1}{y_2} \)
We need to find:
\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{y_1 + y_2}{y_1y_2} \]
Substituting the values from Vieta's:
\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{-\frac{2\frac{a^4k}{hb^2}}{1 + \frac{a^4k^2}{b^4}}}{\frac{\frac{a^4}{h^2} - a^2}{1 + \frac{a^4k^2}{b^4}}} \]
This simplifies to:
\[ \frac{-2\frac{a^4k}{hb^2}}{\frac{a^4}{h^2} - a^2} \]

### Step 8: Final Simplification
After simplification, we find:
\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{2k}{b^2} \cdot \frac{h^2}{a^4 - a^2h^2} \]
This leads to the final result:
\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{2}{k} \]

### Final Answer
Thus, the value of \( \frac{1}{y_1} + \frac{1}{y_2} \) is:
\[ \boxed{\frac{2}{k}} \]