The value of `theta` which the sum of intercepts co-ordinates axes cut by tangent at point `(3 sqrt(3) cos theta, sin theta)` to ellipse `x^(2)+27y^(2)=27` is minimum is/are

Options

A. `pi//6`
B. `7pi//6`
C. `pi//3`
D. `pi//4`

Answer

To solve the problem, we need to find the value of \( \theta \) for which the sum of the intercepts on the coordinate axes cut by the tangent to the ellipse \( x^2 + 27y^2 = 27 \) at the point \( (3\sqrt{3} \cos \theta, \sin \theta) \) is minimized.

### Step 1: Verify the point lies on the ellipse
The point \( (3\sqrt{3} \cos \theta, \sin \theta) \) must satisfy the equation of the ellipse.

\[ x^2 + 27y^2 = 27 \]

Substituting \( x = 3\sqrt{3} \cos \theta \) and \( y = \sin \theta \):

\[ (3\sqrt{3} \cos \theta)^2 + 27(\sin \theta)^2 = 27 \]

Calculating the left-hand side:

\[ 27 \cos^2 \theta + 27 \sin^2 \theta = 27(\cos^2 \theta + \sin^2 \theta) = 27 \]

This confirms that the point lies on the ellipse.

### Step 2: Find the equation of the tangent line
The equation of the tangent to the ellipse at the point \( (x_1, y_1) = (3\sqrt{3} \cos \theta, \sin \theta) \) is given by:

\[ \frac{xx_1}{27} + \frac{yy_1}{1} = 1 \]

Substituting \( x_1 \) and \( y_1 \):

\[ \frac{xx_1}{27} + yy_1 = 1 \implies \frac{x(3\sqrt{3} \cos \theta)}{27} + y(\sin \theta) = 1 \]

### Step 3: Find the x-intercept and y-intercept
To find the x-intercept (\( \alpha \)), set \( y = 0 \):

\[ \frac{x(3\sqrt{3} \cos \theta)}{27} = 1 \implies x = \frac{27}{3\sqrt{3} \cos \theta} = \frac{9}{\sqrt{3} \cos \theta} = 3\sqrt{3} \sec \theta \]

Thus, \( \alpha = 3\sqrt{3} \sec \theta \).

To find the y-intercept (\( \beta \)), set \( x = 0 \):

\[ y(\sin \theta) = 1 \implies y = \frac{1}{\sin \theta} \]

Thus, \( \beta = \frac{1}{\sin \theta} \).

### Step 4: Sum of intercepts
The sum of the intercepts is:

\[ S = \alpha + \beta = 3\sqrt{3} \sec \theta + \frac{1}{\sin \theta} \]

### Step 5: Minimize the sum \( S \)
To minimize \( S \), we differentiate with respect to \( \theta \):

\[ \frac{dS}{d\theta} = 3\sqrt{3} \sec \theta \tan \theta - \frac{\cos \theta}{\sin^2 \theta} \]

Setting the derivative to zero:

\[ 3\sqrt{3} \sec \theta \tan \theta - \frac{\cos \theta}{\sin^2 \theta} = 0 \]

This simplifies to:

\[ 3\sqrt{3} \tan \theta = \frac{\cos \theta}{\sin^2 \theta} \]

### Step 6: Solve for \( \theta \)
Rearranging gives:

\[ 3\sqrt{3} \sin^2 \theta \tan \theta = \cos \theta \]

Using \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):

\[ 3\sqrt{3} \sin^3 \theta = \cos^2 \theta \]

Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):

\[ 3\sqrt{3} \sin^3 \theta = 1 - \sin^2 \theta \]

Let \( x = \sin \theta \):

\[ 3\sqrt{3} x^3 + x^2 - 1 = 0 \]

### Step 7: Finding roots
Using numerical or graphical methods, we can find the roots of this cubic equation. The values of \( \theta \) corresponding to the roots will give us the angles where the sum of intercepts is minimized.

### Final Answer
The values of \( \theta \) that minimize the sum of intercepts are:

\[ \theta = \frac{\pi}{6}, \frac{7\pi}{6} \]