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

Answer

To solve the problem step by step, we will follow the given approach in the video transcript.

### Step-by-Step Solution:

1. **Identify the Center of the Ellipse**:
The center \( C \) of the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is at the point \( C(0, 0) \).

**Hint**: The center of an ellipse in standard form is always at the origin unless otherwise stated.

2. **Determine the Coordinates of the End of the Latus Rectum**:
The coordinates of the end of the latus rectum for the ellipse are given by \( (ae, \frac{b^2}{a}) \), where \( e \) is the eccentricity of the ellipse. The eccentricity \( e \) is defined as \( e = \sqrt{1 - \frac{b^2}{a^2}} \).

**Hint**: Remember that the latus rectum is a line segment perpendicular to the major axis through the foci of the ellipse.

3. **Write the Normal Equation**:
The normal line at a point \( (x_1, y_1) \) on the ellipse is given by the equation:
\[ \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 \]
Substituting \( (x_1, y_1) = (ae, \frac{b^2}{a}) \):
\[ \frac{a^2 x}{ae} - \frac{b^2 y}{\frac{b^2}{a}} = a^2 - b^2 \]
Simplifying this gives:
\[ \frac{a}{e} x - a y = a^2 - b^2 \]

**Hint**: The normal line is derived from the slope of the tangent at the point of interest.

4. **Find the x-intercept (Point G)**:
To find the x-intercept (where \( y = 0 \)):
\[ \frac{a}{e} x = a^2 - b^2 \]
Thus,
\[ x = \frac{e(a^2 - b^2)}{a} \]
Therefore, the coordinates of point \( G \) are \( \left(\frac{e(a^2 - b^2)}{a}, 0\right) \).

**Hint**: Setting \( y = 0 \ in the normal equation helps find the x-intercept. 5. **Calculate the Distance CG**: The distance \( CG \) is the distance from the center \( C(0, 0) \) to the point \( G\left(\frac{e(a^2 - b^2)}{a}, 0\right) \):
\[ CG = \left| \frac{e(a^2 - b^2)}{a} - 0 \right| = \frac{e(a^2 - b^2)}{a} \]

**Hint**: The distance from the origin to any point \( (x, 0) \) is simply the absolute value of \( x \).

6. **Substituting for \( a^2 - b^2 \)**:
We know that \( a^2 - b^2 = a^2 e^2 \) (from the definition of eccentricity). Thus, substituting gives:
\[ CG = \frac{e(a^2 e^2)}{a} = ae^3 \]

**Hint**: Use the relationship between eccentricity and the semi-major and semi-minor axes to simplify expressions.

### Final Answer:
The value of \( CG \) is \( ae^3 \).