`x^2/27+y^2=1`tangent is drawn at `(3sqrt3 costheta,sintheta)`. If tangent meet x-axis and y-axis at A and B . Minimumvalue of sum of intercepts is at `theta`

Options

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

Answer

To solve the problem, we need to find the minimum value of the sum of the intercepts of the tangent drawn to the ellipse given by the equation \( \frac{x^2}{27} + y^2 = 1 \) at the point \( (3\sqrt{3} \cos \theta, \sin \theta) \).

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

### Step 2: Rearranging the tangent equation
Rearranging gives:
\[ \frac{\sqrt{3} x \cos \theta}{9} + y \sin \theta = 1 \]

### Step 3: Find the x-intercept and y-intercept
To find the x-intercept (where \( y = 0 \)):
\[ \frac{\sqrt{3} x \cos \theta}{9} = 1 \implies x = \frac{9}{\sqrt{3} \cos \theta} = 3\sqrt{3} \sec \theta \]
Thus, the x-intercept \( OA = 3\sqrt{3} \sec \theta \).

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

### Step 4: Sum of the intercepts
The sum of the intercepts \( OA + OB \) is:
\[ S(\theta) = 3\sqrt{3} \sec \theta + \frac{1}{\sin \theta} \]

### Step 5: Simplifying the expression
Using the identity \( \sec \theta = \frac{1}{\cos \theta} \) and \( \sin \theta = \frac{1}{\csc \theta} \):
\[ S(\theta) = 3\sqrt{3} \frac{1}{\cos \theta} + \csc \theta \]

### Step 6: Finding the minimum value
To find the minimum value of \( S(\theta) \), we can differentiate \( S(\theta) \) with respect to \( \theta \) and set the derivative to zero:
\[ S'(\theta) = 3\sqrt{3} \sec \theta \tan \theta - \csc \theta \cot \theta \]
Setting \( S'(\theta) = 0 \) gives:
\[ 3\sqrt{3} \sec \theta \tan \theta = \csc \theta \cot \theta \]

### Step 7: Solving for \( \theta \)
From the above equation, we can derive:
\[ \tan^3 \theta = \frac{1}{3\sqrt{3}} \implies \tan \theta = \frac{1}{\sqrt{3}} \]
Thus, \( \theta = \frac{\pi}{6} \) or \( \theta = \frac{5\pi}{6} \).

### Step 8: Conclusion
The minimum value of the sum of the intercepts occurs at \( \theta = \frac{\pi}{6} \).