A variable parabola `y^(2) = 4ax, a` (where `a ne -(1)/(4))` being the parameter, meets the curve `y^(2) +x -y- 2 = 0` at two points. The locus of the point of intersecion of tangents at these points is

Options

A. `x - 2y - 4 = 0`
B. `x - 4y +2 = 0`
C. `x - 4y -1 = 0`
D. `2x - y +1 = 0`

Explanation

To find the locus of the point of intersection of the tangents at the points where the variable parabola \( y^2 = 4ax \) meets the curve \( y^2 + x - y - 2 = 0 \), we can follow these steps:

### Step 1: Write down the equations
The equations we are dealing with are:
1. The parabola: \( y^2 = 4ax \) (Equation 1)
2. The curve: \( y^2 + x - y - 2 = 0 \) (Equation 2)

### Step 2: Substitute the parabola into the curve
From Equation 1, we can express \( x \) in terms of \( y \):
\[ x = \frac{y^2}{4a} \]
Now substitute this expression for \( x \) into Equation 2:
\[ y^2 + \frac{y^2}{4a} - y - 2 = 0 \]
Multiply through by \( 4a \) to eliminate the fraction:
\[ 4ay^2 + y^2 - 4ay - 8a = 0 \]
Combine like terms:
\[ (4a + 1)y^2 - 4ay - 8a = 0 \]

### Step 3: Use the quadratic formula
This is a quadratic equation in \( y \). The roots \( y_1 \) and \( y_2 \) can be found using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4a + 1 \), \( b = -4a \), and \( c = -8a \). Thus,
\[ y = \frac{4a \pm \sqrt{(-4a)^2 - 4(4a + 1)(-8a)}}{2(4a + 1)} \]
Calculating the discriminant:
\[ b^2 - 4ac = 16a^2 + 32a(4a + 1) = 16a^2 + 128a^2 + 32a = 144a^2 + 32a \]
So, the roots become:
\[ y = \frac{4a \pm \sqrt{144a^2 + 32a}}{2(4a + 1)} \]

### Step 4: Find the points of intersection
Let \( t_1 \) and \( t_2 \) be the parameters corresponding to the points of intersection. The coordinates of the points of intersection are:
\[ P_1 = (a t_1^2, 2a t_1), \quad P_2 = (a t_2^2, 2a t_2) \]

### Step 5: Find the point of intersection of tangents
The point of intersection of the tangents at these points can be expressed as:
\[ X = a t_1 t_2, \quad Y = a(t_1 + t_2) \]
Using the relations:
\[ t_1 + t_2 = \frac{4a}{4a + 1}, \quad t_1 t_2 = \frac{-8a}{4a + 1} \]
we can substitute these into the coordinates:
\[ X = a \left(\frac{-8a}{4a + 1}\right), \quad Y = a \left(\frac{4a}{4a + 1}\right) \]

### Step 6: Eliminate the parameter \( a \)
To find the locus, we eliminate \( a \) from the equations for \( X \) and \( Y \):
\[ X = \frac{-8a^2}{4a + 1}, \quad Y = \frac{4a^2}{4a + 1} \]
From \( Y \), we can express \( a \):
\[ a = \frac{Y(4a + 1)}{4} \implies 4a = \frac{4Y}{Y} - 1 \implies a = \frac{Y}{4} + \frac{1}{4} \]
Substituting this back into the equation for \( X \):
\[ X = -2Y + 2 \implies X + 2Y - 2 = 0 \]

### Final Answer
The locus of the point of intersection of the tangents at the points of intersection is:
\[ \boxed{x - 4y + 2 = 0} \]