The line `2x+y=3` cuts the ellipse `4x^(2)+y^(2)=5` at points P and Q. If `theta` is the acute angle between the normals at P and Q, then `theta` is equal to

Options

A. `tan^(-1)((5)/(3))`
B. `sin^(-1)((3)/(sqrt(34)))`
C. `cos^(-1)((3)/(sqrt(34)))`
D. `cot^(-1)((3)/(4))`

Answer

To find the acute angle \( \theta \) between the normals at points \( P \) and \( Q \) where the line \( 2x + y = 3 \) intersects the ellipse \( 4x^2 + y^2 = 5 \), we can follow these steps:

### Step 1: Find the points of intersection \( P \) and \( Q \)

We have the line:
\[ 2x + y = 3 \quad \text{(1)} \]
and the ellipse:
\[ 4x^2 + y^2 = 5 \quad \text{(2)} \]

From equation (1), we can express \( y \) in terms of \( x \):
\[ y = 3 - 2x \]

Substituting this into equation (2):
\[ 4x^2 + (3 - 2x)^2 = 5 \]

Expanding the equation:
\[ 4x^2 + (9 - 12x + 4x^2) = 5 \]
\[ 8x^2 - 12x + 9 - 5 = 0 \]
\[ 8x^2 - 12x + 4 = 0 \]

Dividing the entire equation by 4:
\[ 2x^2 - 3x + 1 = 0 \]

### Step 2: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
\[ x = \frac{3 \pm \sqrt{9 - 8}}{4} \]
\[ x = \frac{3 \pm 1}{4} \]

Calculating the two possible values of \( x \):
1. \( x = \frac{4}{4} = 1 \)
2. \( x = \frac{2}{4} = \frac{1}{2} \)

### Step 3: Find corresponding \( y \) values

Substituting back to find \( y \):
1. For \( x = 1 \):
\[ y = 3 - 2(1) = 1 \quad \Rightarrow \quad P(1, 1) \]
2. For \( x = \frac{1}{2} \):
\[ y = 3 - 2\left(\frac{1}{2}\right) = 2 \quad \Rightarrow \quad Q\left(\frac{1}{2}, 2\right) \]

### Step 4: Find the slopes of the normals at \( P \) and \( Q \)

The equation of the ellipse is:
\[ 4x^2 + y^2 = 5 \]

The slope of the tangent line at any point \( (x_1, y_1) \) on the ellipse can be found using implicit differentiation:
\[ \frac{d}{dx}(4x^2 + y^2) = 0 \implies 8x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{8x}{2y} = -\frac{4x}{y} \]

#### At point \( P(1, 1) \):
\[ \text{slope of tangent} = -\frac{4(1)}{1} = -4 \implies \text{slope of normal} = \frac{1}{4} \]

#### At point \( Q\left(\frac{1}{2}, 2\right) \):
\[ \text{slope of tangent} = -\frac{4\left(\frac{1}{2}\right)}{2} = -1 \implies \text{slope of normal} = 1 \]

### Step 5: Calculate the angle \( \theta \) between the normals

Let \( m_1 = \frac{1}{4} \) and \( m_2 = 1 \). The formula for the angle between two lines with slopes \( m_1 \) and \( m_2 \) is:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting the values:
\[ \tan \theta = \left| \frac{\frac{1}{4} - 1}{1 + \frac{1}{4} \cdot 1} \right| = \left| \frac{-\frac{3}{4}}{1 + \frac{1}{4}} \right| = \left| \frac{-\frac{3}{4}}{\frac{5}{4}} \right| = \frac{3}{5} \]

### Step 6: Find \( \theta \)

Thus, we have:
\[ \theta = \tan^{-1}\left(\frac{3}{5}\right) \]

### Final Answer
The acute angle \( \theta \) between the normals at points \( P \) and \( Q \) is:
\[ \theta = \tan^{-1}\left(\frac{3}{5}\right) \]