Find the equation of tangent at the point `theta=(pi)/(3)` to the ellipse `(x^(2))/(9)+(y^(2))/(4) = 1`

Answer

To find the equation of the tangent at the point where \(\theta = \frac{\pi}{3}\) to the ellipse given by

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

we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse can be rewritten in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where:
- \(a^2 = 9 \Rightarrow a = 3\)
- \(b^2 = 4 \Rightarrow b = 2\)

### Step 2: Find the coordinates of the point on the ellipse
Using the parametric equations for the ellipse, the coordinates of the point corresponding to \(\theta\) are given by:
\[ x_1 = a \cos \theta = 3 \cos \left(\frac{\pi}{3}\right) \]
\[ y_1 = b \sin \theta = 2 \sin \left(\frac{\pi}{3}\right) \]

Calculating these:
- \(\cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \Rightarrow x_1 = 3 \cdot \frac{1}{2} = \frac{3}{2}\)
- \(\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \Rightarrow y_1 = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}\)

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

### Step 3: Write the equation of the tangent line
The equation of the tangent line at the point \((x_1, y_1)\) on the ellipse is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1 = 0 \]

Substituting \(x_1 = \frac{3}{2}\), \(y_1 = \sqrt{3}\), \(a^2 = 9\), and \(b^2 = 4\):
\[ \frac{x \cdot \frac{3}{2}}{9} + \frac{y \cdot \sqrt{3}}{4} - 1 = 0 \]

### Step 4: Simplify the equation
Multiplying through by 36 (the least common multiple of the denominators):
\[ 4 \cdot 3x + 9y \sqrt{3} - 36 = 0 \]
This simplifies to:
\[ 12x + 9y \sqrt{3} - 36 = 0 \]

### Final Equation
Rearranging gives the final equation of the tangent:
\[ 12x + 9y \sqrt{3} = 36 \]