A tangent is drawn to the ellipse `(x^(2))/(27)+y^(2)=1` at the point `(3 sqrt(3) cos theta sin theta)` where `0 lt theta lt (pi)/(2)`. The sum of intecepts of the tangents with the coordinates axes is least when `theta` equals

Options

A. `(pi)/(6)`
B. `(pi)/(3)`
C. `(pi)/(8)`
D. `(pi)/(4)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given ellipse and the point on it
The equation of the ellipse is given by:
\[ \frac{x^2}{27} + y^2 = 1 \]
The point on the ellipse is given as:
\[ (3\sqrt{3} \cos \theta, \sin \theta) \]
where \(0 < \theta < \frac{\pi}{2}\).

### Step 2: Write the equation of the tangent line at the given point
The general formula for the equation of the tangent to the ellipse at a point \((x_0, y_0)\) is:
\[ \frac{xx_0}{27} + yy_0 = 1 \]
Substituting \(x_0 = 3\sqrt{3} \cos \theta\) and \(y_0 = \sin \theta\), we get:
\[ \frac{xx_0}{27} + yy_0 = 1 \implies \frac{x(3\sqrt{3} \cos \theta)}{27} + y(\sin \theta) = 1 \]

### Step 3: Find the x-intercept and y-intercept of the tangent line
To find the x-intercept, set \(y = 0\):
\[ \frac{x(3\sqrt{3} \cos \theta)}{27} = 1 \implies x = \frac{27}{3\sqrt{3} \cos \theta} = \frac{9}{\sqrt{3} \cos \theta} \]

To find the y-intercept, set \(x = 0\):
\[ y(\sin \theta) = 1 \implies y = \frac{1}{\sin \theta} \]

### Step 4: Calculate the sum of the intercepts
The sum of the intercepts \(S\) is given by:
\[ S = \frac{9}{\sqrt{3} \cos \theta} + \frac{1}{\sin \theta} \]

### Step 5: Simplify the expression for S
We can rewrite \(S\) as:
\[ S = 3\sqrt{3} \sec \theta + \csc \theta \]

### Step 6: Differentiate S with respect to \(\theta\) and find critical points
To find the minimum value of \(S\), we differentiate \(S\) with respect to \(\theta\):
\[ \frac{dS}{d\theta} = 3\sqrt{3} \sec \theta \tan \theta - \csc \theta \cot \theta \]
Setting \(\frac{dS}{d\theta} = 0\) gives:
\[ 3\sqrt{3} \sec \theta \tan \theta = \csc \theta \cot \theta \]

### Step 7: Solve for \(\theta\)
Using the identities \(\sec \theta = \frac{1}{\cos \theta}\), \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), \(\csc \theta = \frac{1}{\sin \theta}\), and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), we simplify:
\[ 3\sqrt{3} \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta} \]
This leads to:
\[ 3\sqrt{3} \sin^2 \theta = \cos^2 \theta \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can solve for \(\theta\).

### Step 8: Find the value of \(\theta\)
After solving, we find:
\[ \tan^3 \theta = \frac{1}{3\sqrt{3}} \implies \theta = \frac{\pi}{6} \]

### Final Answer
Thus, the value of \(\theta\) for which the sum of intercepts is least is:
\[ \theta = \frac{\pi}{6} \]