The equation of normal to the ellipse `x^(2)+4y^(2)=9` at the point wherr ithe eccentric angle is `pi//4` is

Options

A. `4x-2y=9`
B. `4 sqrt(2)x-2 sqrt(2)y=9`
C. `4 sqrt(2x)-- sqrt(2 y)= 9`
D. none of these

Answer

To find the equation of the normal to the ellipse \( x^2 + 4y^2 = 9 \) at the point where the eccentric angle is \( \frac{\pi}{4} \), we can follow these steps:

### Step 1: Write the equation of the ellipse in standard form
The given ellipse is \( x^2 + 4y^2 = 9 \). We can rewrite this in standard form by dividing the entire equation by 9:

\[ \frac{x^2}{9} + \frac{4y^2}{9} = 1 \]

This gives us:

\[ \frac{x^2}{3^2} + \frac{y^2}{\left(\frac{3}{2}\right)^2} = 1 \]

From this, we can identify \( a^2 = 9 \) and \( b^2 = \frac{9}{4} \), which leads to \( a = 3 \) and \( b = \frac{3}{2} \).

### Step 2: Find the coordinates of the point on the ellipse
The eccentric angle \( \theta \) is given as \( \frac{\pi}{4} \). The coordinates \( (x, y) \) on the ellipse can be found using the parametric equations:

\[ x = a \cos \theta = 3 \cos\left(\frac{\pi}{4}\right) \]
\[ y = b \sin \theta = \frac{3}{2} \sin\left(\frac{\pi}{4}\right) \]

Calculating these values:

\[ x = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]
\[ y = \frac{3}{2} \cdot \frac{1}{\sqrt{2}} = \frac{3\sqrt{2}}{4} \]

Thus, the point on the ellipse is \( \left(\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{4}\right) \).

### Step 3: Write the equation of the normal line
The equation of the normal to the ellipse in terms of the eccentric angle \( \theta \) is given by:

\[ a x \sec \theta - b y \cos \theta = a^2 - b^2 \]

Substituting \( a = 3 \), \( b = \frac{3}{2} \), and \( \theta = \frac{\pi}{4} \):

\[ 3 x \sec\left(\frac{\pi}{4}\right) - \frac{3}{2} y \cos\left(\frac{\pi}{4}\right) = 9 - \frac{9}{4} \]

Calculating \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):

\[ 3 x \sqrt{2} - \frac{3}{2} y \cdot \frac{1}{\sqrt{2}} = 9 - \frac{9}{4} \]

Calculating the right side:

\[ 9 - \frac{9}{4} = \frac{36}{4} - \frac{9}{4} = \frac{27}{4} \]

Thus, we have:

\[ 3\sqrt{2} x - \frac{3}{2\sqrt{2}} y = \frac{27}{4} \]

### Step 4: Simplify the equation
To eliminate the fractions, we can multiply through by 4:

\[ 12\sqrt{2} x - 6y = 27 \]

Rearranging gives us the equation of the normal:

\[ 12\sqrt{2} x - 6y - 27 = 0 \]

### Final Answer
The equation of the normal to the ellipse at the given point is:

\[ 12\sqrt{2} x - 6y = 27 \]