The tangent and normal to the ellipse `x^(2)+4y^(2)=4` at a point `P(theta)` on it meet the major aixs in Q and R repectively. IF QR=2, then value of `cos theta` is/are

Options

A. `3//4`
B. `-3//4`
C. `2//3`
D. `-2//3`

Answer

To solve the problem, we need to find the value of \( \cos \theta \) given that the tangent and normal to the ellipse \( x^2 + 4y^2 = 4 \) at a point \( P(\theta) \) meet the major axis at points \( Q \) and \( R \) respectively, and that the distance \( QR = 2 \).

### Step 1: Identify the ellipse and its parameters
The given ellipse is \( x^2 + 4y^2 = 4 \). We can rewrite it in standard form:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]
From this, we identify \( a^2 = 4 \) (thus \( a = 2 \)) and \( b^2 = 1 \) (thus \( b = 1 \)). The major axis is along the x-axis.

### Step 2: Parametrize the point \( P(\theta) \)
The point \( P(\theta) \) on the ellipse can be represented as:
\[ P(\theta) = (2 \cos \theta, \sin \theta) \]

### Step 3: Find the equation of the tangent at point \( P(\theta) \)
The equation of the tangent to the ellipse at point \( P(\theta) \) is given by:
\[ \frac{x \cos \theta}{2} + \frac{y \sin \theta}{1} = 1 \]
This simplifies to:
\[ x \cos \theta + 2y \sin \theta = 2 \]

### Step 4: Find the intersection point \( Q \) on the major axis
To find the coordinates of point \( Q \), we set \( y = 0 \) in the tangent equation:
\[ x \cos \theta = 2 \implies x = \frac{2}{\cos \theta} \]
Thus, the coordinates of \( Q \) are:
\[ Q = \left(\frac{2}{\cos \theta}, 0\right) \]

### Step 5: Find the equation of the normal at point \( P(\theta) \)
The slope of the tangent is given by:
\[ -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{\cos \theta}{2 \sin \theta} \]
Thus, the slope of the normal is:
\[ \frac{2 \sin \theta}{\cos \theta} \]
The equation of the normal at \( P(\theta) \) is:
\[ y - \sin \theta = \frac{2 \sin \theta}{\cos \theta}(x - 2 \cos \theta) \]

### Step 6: Find the intersection point \( R \) on the major axis
Setting \( y = 0 \) in the normal equation:
\[ 0 - \sin \theta = \frac{2 \sin \theta}{\cos \theta}(x - 2 \cos \theta) \]
This simplifies to:
\[ -\sin \theta = \frac{2 \sin \theta}{\cos \theta}(x - 2 \cos \theta) \]
Assuming \( \sin \theta \neq 0 \):
\[ -1 = \frac{2}{\cos \theta}(x - 2 \cos \theta) \]
Multiplying through by \( \cos \theta \):
\[ -\cos \theta = 2x - 4 \implies 2x = 4 - \cos \theta \implies x = \frac{4 - \cos \theta}{2} \]
Thus, the coordinates of \( R \) are:
\[ R = \left(\frac{4 - \cos \theta}{2}, 0\right) \]

### Step 7: Calculate the distance \( QR \)
The distance \( QR \) is:
\[ QR = \left| \frac{4 - \cos \theta}{2} - \frac{2}{\cos \theta} \right| = 2 \]
Setting up the equation:
\[ \left| \frac{4 - \cos \theta - 4/\cos \theta}{2} \right| = 2 \]
This leads to:
\[ \left| 4 - \cos \theta - \frac{4}{\cos \theta} \right| = 4 \]

### Step 8: Solve the equation
We can solve the two cases:
1. \( 4 - \cos \theta - \frac{4}{\cos \theta} = 4 \)
2. \( 4 - \cos \theta - \frac{4}{\cos \theta} = -4 \)

From the first case:
\[ -\cos \theta - \frac{4}{\cos \theta} = 0 \implies \cos^2 \theta = -4 \text{ (not possible)} \]

From the second case:
\[ -\cos \theta - \frac{4}{\cos \theta} = -8 \implies \cos \theta + \frac{4}{\cos \theta} = 8 \]
Multiplying by \( \cos \theta \):
\[ \cos^2 \theta - 8 \cos \theta + 4 = 0 \]
Using the quadratic formula:
\[ \cos \theta = \frac{8 \pm \sqrt{64 - 16}}{2} = \frac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3} \]

### Step 9: Determine valid values of \( \cos \theta \)
Since \( \cos \theta \) must be in the range \([-1, 1]\), we check:
\[ 4 - 2\sqrt{3} \text{ and } 4 + 2\sqrt{3} \]
Only \( 4 - 2\sqrt{3} \) is valid.

### Final Answer
Thus, the value of \( \cos \theta \) is:
\[ \cos \theta = 4 - 2\sqrt{3} \]