The angle between the tangents drawn from the origin to the parabola `y^2=4a(x-a)` is

Options

A. `90^@`
B. `30^@`
C. `tan^-1(1/2)`
D. `45^@`

Answer

To find the angle between the tangents drawn from the origin to the parabola given by the equation \(y^2 = 4a(x - a)\), we can follow these steps:

### Step 1: Write the equation of the tangent line
Since the tangents are drawn from the origin (0, 0), we can express the equation of the tangent line in slope-intercept form as:
\[ y = mx \]
where \(m\) is the slope of the tangent.

### Step 2: Substitute the tangent equation into the parabola equation
We substitute \(y = mx\) into the parabola equation \(y^2 = 4a(x - a)\):
\[ (mx)^2 = 4a(x - a) \]
This simplifies to:
\[ m^2x^2 = 4ax - 4a^2 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ m^2x^2 - 4ax + 4a^2 = 0 \]
This is a quadratic equation in \(x\).

### Step 4: Determine the condition for tangency
For the line to be a tangent to the parabola, the quadratic must have exactly one solution. This occurs when the discriminant \(D\) is zero:
\[ D = b^2 - 4ac \]
Here, \(a = m^2\), \(b = -4a\), and \(c = 4a^2\). Thus, we calculate:
\[ D = (-4a)^2 - 4(m^2)(4a^2) = 16a^2 - 16a^2m^2 \]
Setting the discriminant to zero gives:
\[ 16a^2(1 - m^2) = 0 \]
Since \(16a^2 \neq 0\) (as \(a\) is a parameter of the parabola), we have:
\[ 1 - m^2 = 0 \implies m^2 = 1 \]
Thus, \(m = \pm 1\).

### Step 5: Find the angle between the tangents
The slopes of the tangents are \(m_1 = 1\) and \(m_2 = -1\). The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting the values of \(m_1\) and \(m_2\):
\[ \tan \theta = \frac{1 - (-1)}{1 + (1)(-1)} = \frac{2}{0} \]
Since division by zero occurs, \(\tan \theta\) is undefined, indicating that \(\theta = 90^\circ\).

### Conclusion
The angle between the tangents drawn from the origin to the parabola \(y^2 = 4a(x - a)\) is:
\[ \theta = 90^\circ \]