Let `P_(1) : y^(2) = 4ax` and `P_(2) : y^(2) =-4ax` be two parabolas and L : y = x be a straight line.
Equation of the tangent at the point on the parabola `P_(1)` where the line L meets the parabola is

Options

A. `x - 2y + 4a = 0`
B. `x + 2y - 4a = 0`
C. `x + 2y - 8a = 0`
D. `x - 2y + 8a = 0`

Answer

To find the equation of the tangent at the point where the line \( L: y = x \) meets the parabola \( P_1: y^2 = 4ax \), we will follow these steps:

### Step 1: Find the points of intersection
To find the points where the line \( y = x \) intersects the parabola \( y^2 = 4ax \), we substitute \( y = x \) into the equation of the parabola.

\[ x^2 = 4ax \]

Rearranging gives us:

\[ x^2 - 4ax = 0 \]

Factoring out \( x \):

\[ x(x - 4a) = 0 \]

This gives us two solutions:

\[ x = 0 \quad \text{or} \quad x = 4a \]

Thus, the points of intersection are \( (0, 0) \) and \( (4a, 4a) \).

### Step 2: Find the tangent line at the point of intersection
We will find the equation of the tangent line at the point \( (4a, 4a) \) on the parabola \( P_1 \).

The formula for the equation of the tangent to the parabola \( y^2 = 4ax \) at the point \( (x_1, y_1) \) is given by:

\[ yy_1 = 2a(x + x_1) \]

Substituting \( x_1 = 4a \) and \( y_1 = 4a \):

\[ y(4a) = 2a(x + 4a) \]

### Step 3: Simplify the tangent equation
Dividing both sides by \( 4a \):

\[ y = \frac{2a}{4a}(x + 4a) \]

This simplifies to:

\[ y = \frac{1}{2}(x + 4a) \]

Multiplying through by 2 gives:

\[ 2y = x + 4a \]

Rearranging this gives us the final equation of the tangent:

\[ x - 2y + 4a = 0 \]

### Final Result
The equation of the tangent at the point on the parabola \( P_1 \) where the line \( L \) meets the parabola is:

\[ x - 2y + 4a = 0 \]

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