The equation of tangent to the curve `x=a sec theta, y =a tan theta" at "theta=(pi)/(6)` is

Options

A. 2x-y=3a
B. 2x+y=3a
C. `2x-y=sqrt(3)a`
D. `2x+y=sqrt(3)a`

Answer

To find the equation of the tangent to the curve defined by the parametric equations \( x = a \sec \theta \) and \( y = a \tan \theta \) at \( \theta = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Find the coordinates \( (x_1, y_1) \) at \( \theta = \frac{\pi}{6} \)

1. Calculate \( x_1 \):
\[ x_1 = a \sec\left(\frac{\pi}{6}\right) = a \cdot \frac{2}{\sqrt{3}} = \frac{2a}{\sqrt{3}} \]

2. Calculate \( y_1 \):
\[ y_1 = a \tan\left(\frac{\pi}{6}\right) = a \cdot \frac{1}{\sqrt{3}} = \frac{a}{\sqrt{3}} \]

### Step 2: Find the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \)

1. Differentiate \( x \):
\[ \frac{dx}{d\theta} = a \sec\theta \tan\theta \]

2. Differentiate \( y \):
\[ \frac{dy}{d\theta} = a \sec^2\theta \]

### Step 3: Find \( \frac{dy}{dx} \)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sec^2\theta}{a \sec\theta \tan\theta} = \frac{\sec\theta}{\tan\theta} \]

### Step 4: Evaluate \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{6} \)

1. Substitute \( \theta = \frac{\pi}{6} \):
\[ \frac{dy}{dx} \bigg|_{\theta = \frac{\pi}{6}} = \frac{\sec\left(\frac{\pi}{6}\right)}{\tan\left(\frac{\pi}{6}\right)} = \frac{\frac{2}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = 2 \]

### Step 5: Write the equation of the tangent line

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{dy}{dx} \), \( (x_1, y_1) = \left(\frac{2a}{\sqrt{3}}, \frac{a}{\sqrt{3}}\right) \):
\[ y - \frac{a}{\sqrt{3}} = 2\left(x - \frac{2a}{\sqrt{3}}\right) \]

### Step 6: Simplify the equation

1. Distributing the right side:
\[ y - \frac{a}{\sqrt{3}} = 2x - \frac{4a}{\sqrt{3}} \]

2. Rearranging gives:
\[ 2x - y - \frac{4a}{\sqrt{3}} + \frac{a}{\sqrt{3}} = 0 \]
\[ 2x - y - \frac{3a}{\sqrt{3}} = 0 \]

Thus, the equation of the tangent line is:
\[ 2x - y = \frac{3a}{\sqrt{3}} \]

### Final Result
The equation of the tangent to the curve at \( \theta = \frac{\pi}{6} \) is:
\[ 2x - y = \sqrt{3}a \]