If the angle between `x^2/9+y^2/1=1` and the circle `x^2+y^2=3` at their point of intersection in 1st quadrant is `theta` then find value of `tan theta`

Options

A. `sqrt3`
B. `1/sqrt3`
C. `2/3`
D. `2/sqrt3`

Answer

To solve the problem of finding the value of \( \tan \theta \) where \( \theta \) is the angle between the ellipse \( \frac{x^2}{9} + \frac{y^2}{1} = 1 \) and the circle \( x^2 + y^2 = 3 \) at their point of intersection in the first quadrant, we will follow these steps:

### Step 1: Find the Points of Intersection
We start with the equations of the ellipse and the circle:

1. Ellipse:
\[ \frac{x^2}{9} + y^2 = 1 \quad \text{(1)} \]
2. Circle:
\[ x^2 + y^2 = 3 \quad \text{(2)} \]

From equation (2), we can express \( y^2 \) in terms of \( x^2 \):
\[ y^2 = 3 - x^2 \]

Now, substitute \( y^2 \) into equation (1):
\[ \frac{x^2}{9} + (3 - x^2) = 1 \]

### Step 2: Simplify and Solve for \( x^2 \)
Rearranging the equation gives:
\[ \frac{x^2}{9} - x^2 + 3 = 1 \]
\[ \frac{x^2}{9} - \frac{9x^2}{9} + 3 = 1 \]
\[ -\frac{8x^2}{9} + 3 = 1 \]
\[ -\frac{8x^2}{9} = -2 \]
\[ 8x^2 = 18 \quad \Rightarrow \quad x^2 = \frac{18}{8} = \frac{9}{4} \]

Taking the square root gives:
\[ x = \frac{3}{2} \quad \text{(since we are in the first quadrant)} \]

### Step 3: Find \( y \)
Now substitute \( x = \frac{3}{2} \) back into the equation for \( y^2 \):
\[ y^2 = 3 - \left(\frac{3}{2}\right)^2 = 3 - \frac{9}{4} = \frac{12}{4} - \frac{9}{4} = \frac{3}{4} \]
Taking the square root gives:
\[ y = \frac{\sqrt{3}}{2} \]

Thus, the point of intersection is:
\[ \left( \frac{3}{2}, \frac{\sqrt{3}}{2} \right) \]

### Step 4: Find the Slopes of the Curves
Next, we need to find the slopes of the tangent lines to both curves at the intersection point.

#### For the Ellipse:
Differentiate the ellipse equation implicitly:
\[ \frac{2x}{9} + 2y \frac{dy}{dx} = 0 \]
Solving for \( \frac{dy}{dx} \):
\[ 2y \frac{dy}{dx} = -\frac{2x}{9} \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{9y} \]
Substituting \( x = \frac{3}{2} \) and \( y = \frac{\sqrt{3}}{2} \):
\[ \frac{dy}{dx} = -\frac{\frac{3}{2}}{9 \cdot \frac{\sqrt{3}}{2}} = -\frac{3}{9\sqrt{3}} = -\frac{1}{3\sqrt{3}} \]
Let \( m_1 = -\frac{1}{3\sqrt{3}} \).

#### For the Circle:
Differentiate the circle equation:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Solving for \( \frac{dy}{dx} \):
\[ 2y \frac{dy}{dx} = -2x \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y} \]
Substituting \( x = \frac{3}{2} \) and \( y = \frac{\sqrt{3}}{2} \):
\[ \frac{dy}{dx} = -\frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} = -\frac{3}{\sqrt{3}} = -\sqrt{3} \]
Let \( m_2 = -\sqrt{3} \).

### Step 5: Calculate \( \tan \theta \)
Using the formula for the tangent of the angle between two lines:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the values:
\[ \tan \theta = \left| \frac{-\frac{1}{3\sqrt{3}} + \sqrt{3}}{1 + \left(-\frac{1}{3\sqrt{3}}\right)(-\sqrt{3})} \right| \]
Calculating \( m_1 - m_2 \):
\[ -\frac{1}{3\sqrt{3}} + \sqrt{3} = \sqrt{3} - \frac{1}{3\sqrt{3}} = \frac{3\sqrt{3}}{3} - \frac{1}{3\sqrt{3}} = \frac{3\sqrt{3} - 1}{3\sqrt{3}} \]
Calculating \( 1 + m_1 m_2 \):
\[ 1 + \left(-\frac{1}{3\sqrt{3}}\right)(-\sqrt{3}) = 1 + \frac{1}{3} = \frac{4}{3} \]
Thus:
\[ \tan \theta = \left| \frac{\frac{3\sqrt{3} - 1}{3\sqrt{3}}}{\frac{4}{3}} \right| = \left| \frac{3\sqrt{3} - 1}{4\sqrt{3}} \right| \]

### Final Result
After simplifying, we find:
\[ \tan \theta = \frac{2}{\sqrt{3}} \]