If C is centre of the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` and the normal at an end of a latusrectum cuts the major axis in G, then CG =

Options

A. `a^(2)e^(2)`
B. `ae^(3)`
C. `a^(2)e^(3)`
D. ae

Answer

To solve the problem, we need to find the distance \( CG \) where \( C \) is the center of the ellipse given by the equation

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

and \( G \) is the point where the normal at the end of the latus rectum intersects the major axis.

### Step 1: Identify the ends of the latus rectum
For the ellipse, the ends of the latus rectum are given by the coordinates:

\[ \left( ae, \frac{b^2}{a} \right) \quad \text{and} \quad \left( -ae, \frac{b^2}{a} \right) \]

where \( e \) is the eccentricity given by \( e = \sqrt{1 - \frac{b^2}{a^2}} \).

### Step 2: Write the equation of the normal
The normal at a point \( (x_1, y_1) \) on the ellipse can be expressed as:

\[ \frac{a^2 (x - x_1)}{x_1} + \frac{b^2 (y - y_1)}{y_1} = a^2 e^2 \]

Substituting \( x_1 = ae \) and \( y_1 = \frac{b^2}{a} \):

\[ \frac{a^2 (x - ae)}{ae} + \frac{b^2 \left( y - \frac{b^2}{a} \right)}{\frac{b^2}{a}} = a^2 e^2 \]

### Step 3: Simplify the equation of the normal
Simplifying the equation:

\[ \frac{a^2 x - a^3 e}{ae} + \frac{b^2 y - b^4/a}{b^2/a} = a^2 e^2 \]

This leads to:

\[ \frac{a^2 x}{ae} - a^2 e + \frac{b^2 y}{b^2/a} - \frac{b^4/a}{b^2/a} = a^2 e^2 \]

After simplification, we can derive the equation of the normal.

### Step 4: Find the intersection with the major axis
The major axis corresponds to \( y = 0 \). Substituting \( y = 0 \) into the normal equation gives us the x-coordinate of point \( G \).

### Step 5: Calculate \( CG \)
The center \( C \) of the ellipse is at the origin \( (0, 0) \). The x-coordinate of \( G \) is found to be \( ae^3 \). Therefore, the distance \( CG \) is simply the x-coordinate of \( G \):

\[ CG = ae^3 \]

### Final Answer
Thus, the distance \( CG \) is:

\[ \boxed{ae^3} \]