From a point P tangents are drawn to the parabola `y^2=4ax`. If the angle between them is `(pi)/(3)` then locus of P is :

Options

A. `y^2-3x^2+10ax=3a^2`
B. `y^2+3x^2-10ax=3a^2`
C. `y^2-3x^2-10ax=3a^2`
D. `y^2+3x^2+10ax=3a^2`

Answer

To find the locus of the point \( P(h, k) \) from which tangents are drawn to the parabola \( y^2 = 4ax \) such that the angle between the tangents is \( \frac{\pi}{3} \), we can follow these steps:

### Step 1: Write the equation of the parabola
The given parabola is \( y^2 = 4ax \).

### Step 2: Set up the point P
Let the point \( P \) be \( (h, k) \).

### Step 3: Use the formula for the pair of tangents
The equation of the pair of tangents drawn from the point \( P(h, k) \) to the parabola \( y^2 = 4ax \) is given by:
\[ SS_1 = T^2 \]
where \( S \) is the equation of the parabola and \( S_1 \) is obtained by substituting the coordinates of point \( P \) into the equation of the parabola.

- \( S = y^2 - 4ax \)
- \( S_1 = k^2 - 4ah \)
- \( T = ky - 2ax + h \)

### Step 4: Substitute into the equation
Substituting into the equation gives:
\[ (y^2 - 4ax)(k^2 - 4ah) = (ky - 2ax + h)^2 \]

### Step 5: Expand both sides
Expanding both sides:
\[ y^2k^2 - 4ah y^2 - 4axk^2 + 16a^2hx = k^2y^2 - 4ahky + h^2 - 4ax(2a) + 4ahx \]

### Step 6: Collect coefficients
To find the coefficients \( a, b, h \) in the general form of the pair of straight lines, we need to identify:
- Coefficient of \( x^2 \): \( 0 \)
- Coefficient of \( y^2 \): \( k^2 - 4ah \)
- Coefficient of \( xy \): \( -4ak \)

### Step 7: Use the angle between tangents
The angle \( \theta \) between the tangents is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
Given \( \theta = \frac{\pi}{3} \), we have:
\[ \tan \frac{\pi}{3} = \sqrt{3} \]
Thus,
\[ \sqrt{3} = \frac{2\sqrt{h^2 - (4a)(k^2 - 4ah)}}{0 + (k^2 - 4ah)} \]

### Step 8: Simplify the equation
Cross-multiplying gives:
\[ \sqrt{3}(k^2 - 4ah) = 2\sqrt{h^2 - 4a(k^2 - 4ah)} \]
Squaring both sides leads to:
\[ 3(k^2 - 4ah)^2 = 4(h^2 - 4a(k^2 - 4ah)) \]

### Step 9: Rearranging and simplifying
Rearranging and simplifying this equation will yield a quadratic in terms of \( k \) and \( h \).

### Step 10: Final locus equation
After simplification, we arrive at the locus of point \( P \):
\[ k^2 - 3h^2 - 10ah = 3a^2 \]
Replacing \( k \) with \( y \) and \( h \) with \( x \), we get:
\[ y^2 - 3x^2 - 10ax = 3a^2 \]

### Conclusion
The locus of the point \( P \) is given by:
\[ y^2 - 3x^2 - 10ax - 3a^2 = 0 \]