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 will follow the instructions provided in the video transcript and derive the necessary equations to find the value of \(\theta\).

### Step 1: Identify the ellipse and the point on it
The given ellipse is:

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

The point on the ellipse is given as \((3\cos\theta, \sin\theta)\).

### Step 2: Find the parameters \(a\) and \(b\)
From the equation of the ellipse, we can identify:

\[ a^2 = 9 \implies a = 3 \]
\[ b^2 = 1 \implies b = 1 \]

### Step 3: Write the equation of the normal
The equation of the normal to the ellipse at the point \((a\cos\theta, b\sin\theta)\) is given by:

\[ ax \sec\theta + by \csc\theta = a^2 - b^2 \]

Substituting the values of \(a\) and \(b\):

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

### Step 4: Rearranging the normal equation
Rearranging gives us:

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

### Step 5: Find the perpendicular distance from the origin to the normal
The perpendicular distance \(d\) from the origin \((0, 0)\) to the line \(Ax + By + C = 0\) is given by:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Here, \(A = 3\sec\theta\), \(B = \csc\theta\), and \(C = -8\). Thus, the distance from the origin is:

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

### Step 6: Set the distance equal to 2
According to the problem, this distance \(d\) is equal to 2:

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

### Step 7: Solve for \(\theta\)
Cross-multiplying gives:

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

Dividing both sides by 2:

\[ 4 = \sqrt{9\sec^2\theta + \csc^2\theta} \]

Squaring both sides:

\[ 16 = 9\sec^2\theta + \csc^2\theta \]

### Step 8: Substitute \(\sec^2\theta\) and \(\csc^2\theta\)
Using the identities \(\sec^2\theta = 1 + \tan^2\theta\) and \(\csc^2\theta = 1 + \cot^2\theta\):

\[ 16 = 9(1 + \tan^2\theta) + (1 + \cot^2\theta) \]

### Step 9: Simplify the equation
This simplifies to:

\[ 16 = 9 + 9\tan^2\theta + 1 + \cot^2\theta \]
\[ 16 = 10 + 9\tan^2\theta + \cot^2\theta \]
\[ 6 = 9\tan^2\theta + \cot^2\theta \]

### Step 10: Use \(\cot^2\theta = \frac{1}{\tan^2\theta}\)
Let \(x = \tan^2\theta\):

\[ 6 = 9x + \frac{1}{x} \]

Multiplying through by \(x\):

\[ 6x = 9x^2 + 1 \]
\[ 9x^2 - 6x + 1 = 0 \]

### Step 11: Solve the quadratic equation
Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} \]
\[ x = \frac{6 \pm \sqrt{36 - 36}}{18} = \frac{6}{18} = \frac{1}{3} \]

### Step 12: Find \(\tan^2\theta\)
Thus, \(\tan^2\theta = \frac{1}{3}\), which gives:

\[ \tan\theta = \frac{1}{\sqrt{3}} \]

### Step 13: Find \(\theta\)
This corresponds to:

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

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

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