An equation of the normal to the ellipse `x^2//a^2+y^2//b^2=1` with eccentricity e at the positive end of the latus rectum is

Options

A. `x+ey+e^3 a=0`
B. `x-ey-ae^3=0`
C. `x-ey+e^3 a=0`
D. `x+ey-e^3 a=0`

Answer

To find the equation of the normal to the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at the positive end of the latus rectum, we can follow these steps:

### Step 1: Identify the Eccentricity and Coordinates of the End of the Latus Rectum
The eccentricity \( e \) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
The coordinates of the positive end of the latus rectum can be found using:
\[ \left( ae, \frac{b^2}{a} \right) \]
Thus, the coordinates are \( \left( ae, \frac{b^2}{a} \right) \).

### Step 2: Write the General Equation of the Normal
The general equation of the normal to the ellipse at a point \( (x_1, y_1) \) is given by:
\[ \frac{a^2}{x_1} (x - x_1) + \frac{b^2}{y_1} (y - y_1) = - (a^2 + b^2) \]
In our case, \( x_1 = ae \) and \( y_1 = \frac{b^2}{a} \).

### Step 3: Substitute the Coordinates into the Normal Equation
Substituting \( x_1 \) and \( y_1 \) into the normal equation:
\[ \frac{a^2}{ae} (x - ae) + \frac{b^2}{\frac{b^2}{a}} (y - \frac{b^2}{a}) = - (a^2 + b^2) \]
This simplifies to:
\[ \frac{a}{e} (x - ae) + a (y - \frac{b^2}{a}) = - (a^2 + b^2) \]

### Step 4: Simplify the Equation
Distributing the terms:
\[ \frac{a}{e} x - \frac{a^2}{e} + ay - b^2 = - (a^2 + b^2) \]
Rearranging gives:
\[ \frac{a}{e} x + ay = - (a^2 + b^2) + \frac{a^2}{e} + b^2 \]
This leads to:
\[ \frac{a}{e} x + ay = \frac{a^2}{e} - a^2 \]

### Step 5: Final Form of the Normal Equation
Rearranging the equation gives:
\[ x - ey = ae^2 \]
Thus, the equation of the normal at the positive end of the latus rectum is:
\[ x - ey - ae^2 = 0 \]

### Conclusion
The equation of the normal to the ellipse at the positive end of the latus rectum is:
\[ x - ey - ae^2 = 0 \]