The angle between tangents drawn to the curve `xy = 4` from the point `(1, 3)` is

Options

A. `30^(@)`
B. `60^(@)`
C. `tan^(-1) . 4/5`
D. `tan^(-1) . 5/4`

Answer

To find the angle between the tangents drawn to the curve \(xy = 4\) from the point \((1, 3)\), we can follow these steps:

### Step 1: Understand the curve and the point
The given curve is \(xy = 4\) and the point from which we are drawing tangents is \((1, 3)\).

### Step 2: Parametric representation of the curve
We can express the curve in parametric form. For the hyperbola \(xy = c^2\), we can use:
\[ x = ct, \quad y = \frac{c^2}{x} = \frac{c}{t} \]
For our case, \(c^2 = 4\) (so \(c = 2\)), the parametric equations become:
\[ x = 2t, \quad y = \frac{4}{2t} = \frac{2}{t} \]

### Step 3: Find the equation of the tangent
The slope of the tangent to the hyperbola at the point \((2t, \frac{2}{t})\) can be derived from the implicit differentiation of \(xy = 4\):
\[ \frac{dy}{dx} = -\frac{y}{x} = -\frac{\frac{2}{t}}{2t} = -\frac{1}{t^2} \]
Thus, the equation of the tangent line at the point \((2t, \frac{2}{t})\) is:
\[ y - \frac{2}{t} = -\frac{1}{t^2}(x - 2t) \]
Rearranging gives:
\[ y = -\frac{1}{t^2}x + \frac{2}{t} + \frac{2}{t} = -\frac{1}{t^2}x + \frac{4}{t} \]

### Step 4: Substitute the point (1, 3) into the tangent equation
To find the values of \(t\) for which the tangents pass through the point \((1, 3)\), we substitute \(x = 1\) and \(y = 3\):
\[ 3 = -\frac{1}{t^2}(1) + \frac{4}{t} \]
Multiplying through by \(t^2\) to eliminate the fraction:
\[ 3t^2 = -1 + 4t \]
Rearranging gives:
\[ 3t^2 - 4t + 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
This gives:
\[ t_1 = 1 \quad \text{and} \quad t_2 = \frac{1}{3} \]

### Step 6: Find the slopes of the tangents
Now, we can find the slopes of the tangents corresponding to \(t_1\) and \(t_2\):
\[ m_1 = -\frac{1}{t_1^2} = -1 \quad \text{and} \quad m_2 = -\frac{1}{t_2^2} = -9 \]

### Step 7: Calculate the angle between the tangents
The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the values:
\[ \tan \theta = \left| \frac{-1 - (-9)}{1 + (-1)(-9)} \right| = \left| \frac{8}{10} \right| = \frac{4}{5} \]

### Step 8: Find the angle \(\theta\)
Thus, the angle \(\theta\) is:
\[ \theta = \tan^{-1}\left(\frac{4}{5}\right) \]

### Final Answer
The angle between the tangents drawn to the curve \(xy = 4\) from the point \((1, 3)\) is \(\tan^{-1}\left(\frac{4}{5}\right)\).

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