A normal is drawn to the ellipse `(x^(2))/(9)+y^(2)=1` at the point `(3cos theta, sin theta)` where `0lt theta lt(pi)/(2)`. If N is the foot of the perpendicular from the origin O to the normal such that ON = 2, then `theta` is equal to

Options

A. `(pi)/(4)`
B. `(pi)/(12)`
C. `(pi)/(3)`
D. `(pi)/(6)`

Answer

To solve the problem step by step, we need to find the angle \( \theta \) for the normal drawn to the ellipse at the point \( (3 \cos \theta, \sin \theta) \) such that the distance from the origin to the foot of the perpendicular from the origin to the normal is 2.

### Step 1: Write the equation of the ellipse
The equation of the ellipse is given by:

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

### Step 2: Identify the point on the ellipse
The point on the ellipse is given as \( (3 \cos \theta, \sin \theta) \).

### Step 3: Write the equation of the normal
The normal to the ellipse at the point \( (3 \cos \theta, \sin \theta) \) can be derived using the formula for the normal to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):

\[ a = 3, \quad b = 1 \]

The equation of the normal at the point \( (a \cos \theta, b \sin \theta) \) is:

\[ a x \sec \theta - b y \csc \theta = a^2 - b^2 \]

Substituting \( a = 3 \) and \( b = 1 \):

\[ 3x \sec \theta - y \csc \theta = 9 - 1 = 8 \]

Thus, the equation of the normal becomes:

\[ 3x \sec \theta - y \csc \theta = 8 \]

### Step 4: Find the distance from the origin to the normal
The distance \( d \) from the origin \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

Rearranging the normal equation:

\[ 3x \sec \theta - y - 8 = 0 \]

Here, \( A = 3 \sec \theta \), \( B = -1 \), and \( C = -8 \).

Thus, the distance from the origin is:

\[ d = \frac{|0 + 0 - 8|}{\sqrt{(3 \sec \theta)^2 + (-1)^2}} = \frac{8}{\sqrt{9 \sec^2 \theta + 1}} \]

### Step 5: Set the distance equal to 2
We know that \( ON = 2 \):

\[ \frac{8}{\sqrt{9 \sec^2 \theta + 1}} = 2 \]

### Step 6: Solve for \( \theta \)
Cross-multiplying gives:

\[ 8 = 2 \sqrt{9 \sec^2 \theta + 1} \]

Squaring both sides:

\[ 64 = 4(9 \sec^2 \theta + 1) \]

Expanding and simplifying:

\[ 64 = 36 \sec^2 \theta + 4 \]
\[ 60 = 36 \sec^2 \theta \]
\[ \sec^2 \theta = \frac{60}{36} = \frac{5}{3} \]

Since \( \sec^2 \theta = 1 + \tan^2 \theta \):

\[ 1 + \tan^2 \theta = \frac{5}{3} \]
\[ \tan^2 \theta = \frac{5}{3} - 1 = \frac{2}{3} \]

Taking the square root:

\[ \tan \theta = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \]

### Step 7: Find \( \theta \)
Now, we can find \( \theta \):

\[ \theta = \tan^{-1}\left(\frac{\sqrt{6}}{3}\right) \]

### Final Result
Thus, the value of \( \theta \) is:

\[ \theta = \frac{\pi}{6} \text{ (or } 30^\circ\text{)} \]