The equation of the normal to the ellipse at the point whose eccentric angle `theta=(pi)/(6)` is

Answer

To find the equation of the normal to the ellipse at the point whose eccentric angle is \(\theta = \frac{\pi}{6}\), we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

### Step 2: Find the coordinates of the point on the ellipse
Using the parametric equations of the ellipse, the coordinates of the point \(P\) corresponding to the eccentric angle \(\theta\) are given by:
\[ P = (a \cos \theta, b \sin \theta) \]
Substituting \(\theta = \frac{\pi}{6}\):
\[ P = \left(a \cos\left(\frac{\pi}{6}\right), b \sin\left(\frac{\pi}{6}\right)\right) \]
Using the known values:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus, the coordinates of point \(P\) become:
\[ P = \left(a \cdot \frac{\sqrt{3}}{2}, b \cdot \frac{1}{2}\right) \]

### Step 3: Write the equation of the normal
The equation of the normal to the ellipse at the point \(P\) can be expressed as:
\[ \frac{a}{\cos \theta}(x - a \cos \theta) - \frac{b}{\sin \theta}(y - b \sin \theta) = 0 \]
Substituting \(\theta = \frac{\pi}{6}\):
\[ \frac{a}{\frac{\sqrt{3}}{2}}(x - a \cdot \frac{\sqrt{3}}{2}) - \frac{b}{\frac{1}{2}}(y - b \cdot \frac{1}{2}) = 0 \]
This simplifies to:
\[ \frac{2a}{\sqrt{3}}(x - \frac{a\sqrt{3}}{2}) - 2b(y - \frac{b}{2}) = 0 \]

### Step 4: Rearranging the equation
Multiplying through by \(\sqrt{3}\) to eliminate the fraction:
\[ 2a(x - \frac{a\sqrt{3}}{2}) - 2b\sqrt{3}(y - \frac{b}{2}) = 0 \]
Expanding this:
\[ 2ax - a^2\sqrt{3} - 2b\sqrt{3}y + b^2\sqrt{3} = 0 \]
Rearranging gives:
\[ 2ax - 2b\sqrt{3}y = a^2\sqrt{3} - b^2\sqrt{3} \]

### Final Equation
Thus, the equation of the normal to the ellipse at the point where the eccentric angle is \(\frac{\pi}{6}\) is:
\[ 2ax - 2b\sqrt{3}y = (a^2 - b^2)\sqrt{3} \]