The tangent and normal to the ellipse `x^2+4y^2=4` at a point `P(theta)` on it meets the major axis in Q and R respectively. If `0 < theta < pi/2` and `QR=2` then show that `theta=`

Options

A. `cos^(-1)(2/3)`
B. `cos^(-1)(1/3)`
C. `cos^(-1)(3/4)`
D. None of these

Answer

To solve the problem, we need to find the angle \( \theta \) such that the length of the segment \( QR \) is equal to 2, where \( Q \) and \( R \) are the points where the tangent and normal to the ellipse \( x^2 + 4y^2 = 4 \) at the point \( P(\theta) \) intersect the major axis.

### Step-by-Step Solution:

1. **Equation of the Ellipse**:
The given equation of the ellipse is:
\[ x^2 + 4y^2 = 4 \]
Dividing both sides by 4, we rewrite it in standard form:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]
This shows that the semi-major axis \( a = 2 \) and the semi-minor axis \( b = 1 \).

2. **Parametric Coordinates**:
The parametric coordinates for the ellipse can be defined as:
\[ x = 2 \cos \theta, \quad y = \sin \theta \]
where \( 0 < \theta < \frac{\pi}{2} \).

3. **Equation of the Tangent**:
The equation of the tangent to the ellipse at the point \( P(\theta) \) is given by:
\[ \frac{x \cos \theta}{2} + y \sin \theta = 1 \]

4. **Finding Point Q**:
To find the intersection of the tangent with the major axis (where \( y = 0 \)):
\[ \frac{x \cos \theta}{2} + 0 \cdot \sin \theta = 1 \implies x \cos \theta = 2 \implies x = \frac{2}{\cos \theta} \]
Thus, the coordinates of point \( Q \) are:
\[ Q\left(\frac{2}{\cos \theta}, 0\right) \]

5. **Equation of the Normal**:
The slope of the normal at point \( P(\theta) \) is the negative reciprocal of the slope of the tangent. The slope of the tangent can be found using implicit differentiation or from the tangent equation. The normal line can be derived as:
\[ y - \sin \theta = -\frac{2 \sin \theta}{\cos \theta} (x - 2 \cos \theta) \]

Rearranging gives:
\[ y = -2 \tan \theta (x - 2 \cos \theta) + \sin \theta \]

6. **Finding Point R**:
Setting \( y = 0 \) for the normal line:
\[ 0 = -2 \tan \theta (x - 2 \cos \theta) + \sin \theta \]
Solving for \( x \):
\[ 2 \tan \theta (x - 2 \cos \theta) = \sin \theta \implies x - 2 \cos \theta = \frac{\sin \theta}{2 \tan \theta} \]
Thus,
\[ x = 2 \cos \theta + \frac{\sin \theta}{2 \tan \theta} = 2 \cos \theta + \frac{\sin^2 \theta}{2 \sin \theta \cos \theta} = 2 \cos \theta + \frac{1}{2 \cos \theta} \]
Therefore, the coordinates of point \( R \) are:
\[ R\left(2 \cos \theta + \frac{1}{2 \cos \theta}, 0\right) \]

7. **Length of Segment QR**:
The distance \( QR \) can be calculated as:
\[ QR = \left(2 \cos \theta + \frac{1}{2 \cos \theta}\right) - \left(\frac{2}{\cos \theta}\right) = 2 \cos \theta + \frac{1}{2 \cos \theta} - \frac{2}{\cos \theta} \]
Simplifying gives:
\[ QR = 2 \cos \theta - \frac{4}{2 \cos \theta} + \frac{1}{2 \cos \theta} = 2 \cos \theta - \frac{3}{2 \cos \theta} \]

8. **Setting QR = 2**:
We set the length \( QR \) equal to 2:
\[ 2 \cos \theta - \frac{3}{2 \cos \theta} = 2 \]
Multiplying through by \( 2 \cos \theta \) to eliminate the fraction:
\[ 4 \cos^2 \theta - 3 = 4 \cos \theta \]
Rearranging gives:
\[ 4 \cos^2 \theta - 4 \cos \theta - 3 = 0 \]

9. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula:
\[ \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \]
Simplifying gives:
\[ \cos \theta = \frac{4 \pm \sqrt{16 + 48}}{8} = \frac{4 \pm \sqrt{64}}{8} = \frac{4 \pm 8}{8} \]
This results in:
\[ \cos \theta = \frac{12}{8} = \frac{3}{2} \quad \text{(not valid)} \quad \text{or} \quad \cos \theta = \frac{-4}{8} = -\frac{1}{2} \quad \text{(not valid)} \]

The valid solution is:
\[ \cos \theta = \frac{2}{3} \]

10. **Finding \( \theta \)**:
Therefore, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]

### Final Answer:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]