The graph of the conic `x^2-(y-1)^2=1` has one tangent line with positive slope that passes through the origin . The point of the tangency being `(a,b)` then find the value of `sin^-1(a/b)`

Options

A. `(5pi)/(12)`
B. `(pi)/(6)`
C. `(pi)/(4)`
D. `(pi)/(3)`

Answer

To solve the problem step by step, we will begin with the given equation of the hyperbola and derive the necessary information to find the required value.

### Step 1: Identify the conic section
The given equation is:
\[ x^2 - (y - 1)^2 = 1 \]
This represents a hyperbola centered at (0, 1) with a transverse axis along the x-axis.

### Step 2: Write the equation of the tangent line
The general equation of the tangent line to a hyperbola in the form \(x^2/a^2 - (y - k)^2/b^2 = 1\) is given by:
\[ y - k = mx \pm \sqrt{a^2m^2 - b^2} \]
Here, \(a = 1\) and \(b = 1\) (from the equation), and \(k = 1\). Thus, the equation becomes:
\[ y - 1 = mx \pm \sqrt{m^2 - 1} \]

### Step 3: Substitute the point (0, 0)
Since the tangent line passes through the origin (0, 0), we substitute \(x = 0\) and \(y = 0\) into the tangent equation:
\[ 0 - 1 = m(0) \pm \sqrt{m^2 - 1} \]
This simplifies to:
\[ -1 = \pm \sqrt{m^2 - 1} \]

### Step 4: Solve for \(m\)
Squaring both sides gives:
\[ 1 = m^2 - 1 \]
Thus,
\[ m^2 = 2 \quad \Rightarrow \quad m = \sqrt{2} \quad (\text{since we want the positive slope}) \]

### Step 5: Find the point of tangency
Now we substitute \(m = \sqrt{2}\) back into the tangent equation:
\[ y - 1 = \sqrt{2}x \pm 1 \]
We will consider the minus sign first:
\[ y - 1 = \sqrt{2}x - 1 \quad \Rightarrow \quad y = \sqrt{2}x \]
Now, we need to find the coordinates of the point of tangency \((a, b)\) that lies on both the hyperbola and the tangent line.

### Step 6: Substitute \(y = \sqrt{2}x\) into the hyperbola equation
Substituting \(y\) into the hyperbola equation:
\[ x^2 - (\sqrt{2}x - 1)^2 = 1 \]
Expanding this gives:
\[ x^2 - (2x^2 - 2\sqrt{2}x + 1) = 1 \]
\[ x^2 - 2x^2 + 2\sqrt{2}x - 1 = 1 \]
\[ -x^2 + 2\sqrt{2}x - 2 = 0 \]
Multiplying through by -1:
\[ x^2 - 2\sqrt{2}x + 2 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \]
\[ = \frac{2\sqrt{2} \pm \sqrt{8 - 8}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \]
Thus, \(a = \sqrt{2}\).

### Step 8: Find \(b\)
Substituting \(a = \sqrt{2}\) back into \(y = \sqrt{2}x\):
\[ b = \sqrt{2}(\sqrt{2}) = 2 \]

### Step 9: Calculate \( \sin^{-1}(a/b) \)
Now we need to find:
\[ \sin^{-1}\left(\frac{a}{b}\right) = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) \]
The value of \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\) is:
\[ \frac{\pi}{4} \]

### Final Answer
Thus, the value of \(\sin^{-1}(a/b)\) is:
\[ \boxed{\frac{\pi}{4}} \]