Equation of a family of circle passing through the extremities of the latus rectum of the parabola `y^(2)=4ax`, g being a parametric is

Options

A. `x^(2)+y^(2)+2g(y-2a)-5a^(2)=0`
B. `x^(2)+y^(2)+2g(x+a)-5a^(2)=0`
C. `x^(2)+y^(2)+2g(x-a)-5a^(2)=0`
D. `x^(2)+y^(2)+2g(y+2a)-5a^(2)=0`

Answer

To find the equation of a family of circles passing through the extremities of the latus rectum of the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Identify the Extremities of the Latus Rectum
The latus rectum of the parabola \( y^2 = 4ax \) is a vertical line segment that passes through the focus of the parabola. The coordinates of the focus are \( (a, 0) \). The endpoints of the latus rectum can be found by substituting \( x = a \) into the parabola's equation:

\[ y^2 = 4a(a) \implies y^2 = 4a^2 \implies y = 2a \text{ or } y = -2a \]

Thus, the extremities of the latus rectum are \( (a, 2a) \) and \( (a, -2a) \).

### Step 2: Use the General Equation of a Circle
The general equation of a circle in the Cartesian plane can be expressed as:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 3: Substitute the Points into the Circle's Equation
We need to ensure that the circle passes through the points \( (a, 2a) \) and \( (a, -2a) \).

1. Substituting \( (a, 2a) \):

\[ a^2 + (2a)^2 + 2ga + 2f(2a) + c = 0 \]
\[ a^2 + 4a^2 + 2ga + 4fa + c = 0 \implies 5a^2 + 2ga + 4fa + c = 0 \tag{1} \]

2. Substituting \( (a, -2a) \):

\[ a^2 + (-2a)^2 + 2ga + 2f(-2a) + c = 0 \]
\[ a^2 + 4a^2 + 2ga - 4fa + c = 0 \implies 5a^2 + 2ga - 4fa + c = 0 \tag{2} \]

### Step 4: Set Up the System of Equations
Now we have two equations (1) and (2):

1. \( 5a^2 + 2ga + 4fa + c = 0 \)
2. \( 5a^2 + 2ga - 4fa + c = 0 \)

### Step 5: Subtract the Equations
Subtract equation (2) from equation (1):

\[ (5a^2 + 2ga + 4fa + c) - (5a^2 + 2ga - 4fa + c) = 0 \]
This simplifies to:

\[ 8fa = 0 \]

Since \( a \neq 0 \), we conclude that \( f = 0 \).

### Step 6: Substitute \( f = 0 \) into One of the Equations
Substituting \( f = 0 \) into equation (1):

\[ 5a^2 + 2ga + c = 0 \implies c = -5a^2 - 2ga \]

### Step 7: Write the Final Equation of the Circle
Now substituting \( f = 0 \) into the general circle equation gives us:

\[ x^2 + y^2 + 2gx - 5a^2 - 2ga = 0 \]

Rearranging gives:

\[ x^2 + y^2 + 2gx - 5a^2 = 2ga \]

This is the equation of the family of circles passing through the extremities of the latus rectum of the parabola \( y^2 = 4ax \).

### Final Answer
The equation of the family of circles is:

\[ x^2 + y^2 + 2gx - 5a^2 = 2ga \]