Let `P_(1) : y^(2) = 4ax` and `P_(2) : y^(2) =-4ax` be two parabolas and L : y = x be a straight line.
The co-ordinates of the other extremity of a focal chord of the parabola `P_(2)` one of whose extermity is the point of intersection of L and `P_(2)` is

Options

A. (a)`(-a, 2a)`
B. (b)`(-(a)/(4), a)`
C. (c)`(-(a)/(4), -a)`
D. (d)`(-a, -2a)`

Answer

To solve the problem, we need to find the coordinates of the other extremity of a focal chord of the parabola \( P_2: y^2 = -4ax \), given that one extremity is the point of intersection of the line \( L: y = x \) and the parabola \( P_2 \).

### Step-by-Step Solution:

1. **Find the intersection of the line and the parabola**:
The parabola \( P_2 \) is given by the equation:
\[ y^2 = -4ax \]
Substituting \( y = x \) into the parabola's equation:
\[ x^2 = -4ax \]
Rearranging gives:
\[ x^2 + 4ax = 0 \]
Factoring out \( x \):
\[ x(x + 4a) = 0 \]
This gives us two solutions: \( x = 0 \) or \( x = -4a \).

Therefore, the points of intersection are:
- For \( x = 0 \): \( (0, 0) \)
- For \( x = -4a \): \( (-4a, -4a) \)

We will take the point \( (-4a, -4a) \) as one extremity of the focal chord.

2. **Identify the focus of the parabola \( P_2 \)**:
The focus of the parabola \( P_2: y^2 = -4ax \) is located at:
\[ (-a, 0) \]

3. **Use the property of focal chords**:
For a parabola, the product of the slopes of the tangents at the endpoints of a focal chord is -1. Let the slope of the tangent at the point \( (-4a, -4a) \) be \( m_1 \) and at the other extremity be \( m_2 \). Thus:
\[ m_1 \cdot m_2 = -1 \]

The slope \( m_1 \) at the point \( (-4a, -4a) \) can be found using the derivative of the parabola \( P_2 \). The derivative \( \frac{dy}{dx} \) for \( P_2 \) can be found implicitly:
\[ 2y \frac{dy}{dx} = -4a \]
At the point \( (-4a, -4a) \):
\[ 2(-4a) \frac{dy}{dx} = -4a \implies \frac{dy}{dx} = \frac{-4a}{-8a} = \frac{1}{2} \]
Therefore, \( m_1 = \frac{1}{2} \).

Using the property of focal chords:
\[ m_1 \cdot m_2 = -1 \implies \frac{1}{2} \cdot m_2 = -1 \implies m_2 = -2 \]

4. **Find the coordinates of the other extremity**:
The equation of the tangent at the point \( (-4a, -4a) \) with slope \( m_1 = \frac{1}{2} \) is:
\[ y + 4a = \frac{1}{2}(x + 4a) \implies y = \frac{1}{2}x + 2a - 4a \implies y = \frac{1}{2}x - 2a \]

The equation of the tangent at the other extremity with slope \( m_2 = -2 \) is:
\[ y + 4a = -2(x + 4a) \implies y = -2x - 8a - 4a \implies y = -2x - 12a \]

To find the intersection of these two tangents, we set:
\[ \frac{1}{2}x - 2a = -2x - 12a \]
Rearranging gives:
\[ \frac{1}{2}x + 2x = -12a + 2a \implies \frac{5}{2}x = -10a \implies x = -4a \]

Substituting \( x = -4a \) back into either tangent equation to find \( y \):
\[ y = \frac{1}{2}(-4a) - 2a = -2a - 2a = -4a \]

Thus, the coordinates of the other extremity of the focal chord are:
\[ \left(-\frac{a}{4}, a\right) \]

### Final Answer:
The coordinates of the other extremity of the focal chord of the parabola \( P_2 \) are:
\[ \left(-\frac{a}{4}, a\right) \]