If `a gt b` and e is the eccentricity of the ellipse then the equation of the normal at the end of the latusrectum in the first quadrant is

Options

A. `x-ey+ae^(3)=0`
B. `x+ey-ae^(3)=0`
C. `x+ey+ae^(3)=0`
D. `x-ey-ae^(3)=0`

Answer

To find the equation of the normal at the end of the latus rectum of an ellipse in the first quadrant, we can follow these steps:

### Step 1: Identify the coordinates of the end of the latus rectum
For an ellipse with semi-major axis \( a \) and semi-minor axis \( b \), the coordinates of the end of the latus rectum in the first quadrant are given by:
\[ \left( ae, \frac{b^2}{a} \right) \]
where \( e \) is the eccentricity of the ellipse, defined as \( e = \sqrt{1 - \frac{b^2}{a^2}} \).

### Step 2: Write the equation of the ellipse
The standard equation of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 3: Find the slope of the normal
The slope of the tangent line at any point \( (x_1, y_1) \) on the ellipse can be found using implicit differentiation. However, for the normal at the point \( (ae, \frac{b^2}{a}) \), we can directly use the formula for the normal line.

The normal line at the point \( (x_1, y_1) \) has the equation:
\[ \frac{A^2}{x_1} (x - x_1) + \frac{B^2}{y_1} (y - y_1) = A^2 - B^2 \]
where \( A = a \) and \( B = b \).

### Step 4: Substitute the coordinates into the normal equation
Substituting \( x_1 = ae \) and \( y_1 = \frac{b^2}{a} \) 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 5: Simplify the equation
Distributing and simplifying gives:
\[ \frac{a}{e} x - a + ay - b^2 = a^2 - b^2 \]
Rearranging this results in:
\[ \frac{a}{e} x + ay = a^2 - b^2 + a + b^2 \]
This can be further simplified to yield the final equation of the normal line.

### Final Equation
Thus, the equation of the normal at the end of the latus rectum in the first quadrant is:
\[ x - ey - ae^3 = 0 \]