The angle between a pair of tangents drawn from a point P to the hyperbola `y^2 = 4ax` is `45^@`. Show that the locus of the point P is hyperbola.

Options

A. `(a, 0)`
B. `(-7a, 0)`
C. `(4a, 0)`
D. `(-4a, 0)`

Answer

To solve the problem step by step, we will derive the locus of the point P from which tangents are drawn to the hyperbola \( y^2 = 4ax \) such that the angle between the tangents is \( 45^\circ \).

### Step 1: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola \( y^2 = 4ax \) at a point can be given as:
\[ y = mx + \frac{a}{m} \]
where \( m \) is the slope of the tangent.

### Step 2: Set the coordinates of point P
Let the coordinates of point P be \( (h, k) \). The equation of the tangent line at point P can be rewritten as:
\[ k = mh + \frac{a}{m} \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ mk - mh - a = 0 \]
This is a quadratic equation in \( m \):
\[ m^2h - mk + a = 0 \]

### Step 4: Identify the roots of the quadratic
Let the roots of this quadratic equation be \( m_1 \) and \( m_2 \). The angle \( \theta \) between the two tangents can be expressed using the slopes:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Given that \( \theta = 45^\circ \), we have:
\[ \tan 45^\circ = 1 \]
Thus, we can set up the equation:
\[ 1 = \frac{m_1 - m_2}{1 + m_1 m_2} \]

### Step 5: Express \( m_1 + m_2 \) and \( m_1 m_2 \)
From Vieta's formulas, we know:
\[ m_1 + m_2 = \frac{k}{h} \quad \text{and} \quad m_1 m_2 = \frac{a}{h} \]

### Step 6: Substitute into the angle equation
Substituting these into the angle equation gives:
\[ 1 + \frac{a}{h} = m_1 - m_2 \]

### Step 7: Find \( m_1 - m_2 \)
Using the identity for the difference of roots:
\[ m_1 - m_2 = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2} \]
Substituting our values:
\[ m_1 - m_2 = \sqrt{\left(\frac{k}{h}\right)^2 - 4\left(\frac{a}{h}\right)} \]

### Step 8: Equate and simplify
Equating the two expressions for \( m_1 - m_2 \):
\[ 1 + \frac{a}{h} = \sqrt{\left(\frac{k}{h}\right)^2 - \frac{4a}{h}} \]
Squaring both sides leads to:
\[ \left(1 + \frac{a}{h}\right)^2 = \left(\frac{k}{h}\right)^2 - \frac{4a}{h} \]

### Step 9: Clear the fractions
Multiply through by \( h^2 \):
\[ (h + a)^2 = k^2 - 4ah \]

### Step 10: Rearranging to find the locus
Rearranging gives:
\[ k^2 = (h + a)^2 + 4ah \]
Expanding the left side:
\[ k^2 = h^2 + 2ah + a^2 + 4ah = h^2 + 6ah + a^2 \]

### Step 11: Final form
This can be rearranged to:
\[ k^2 - h^2 - 6ah - a^2 = 0 \]
This is the equation of a hyperbola.

### Conclusion
Thus, we have shown that the locus of the point P is a hyperbola.

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