If the tangent to the ellipse `x^2 +4y^2=16` at the point 0 sanormal to the circle `x^2 +y^2-8x-4y=0` then `theta` is equal to

Options

A. `pi/2`
B. `pi/4`
C. 0
D. `-pi/4`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the equations
We have the ellipse given by the equation:
\[ x^2 + 4y^2 = 16 \]
We can rewrite this in standard form:
\[ \frac{x^2}{16} + \frac{y^2}{4} = 1 \]
This indicates that \(a^2 = 16\) and \(b^2 = 4\), hence \(a = 4\) and \(b = 2\).

The circle is given by the equation:
\[ x^2 + y^2 - 8x - 4y = 0 \]
We can complete the square for this equation:
\[ (x^2 - 8x) + (y^2 - 4y) = 0 \]
\[ (x - 4)^2 - 16 + (y - 2)^2 - 4 = 0 \]
\[ (x - 4)^2 + (y - 2)^2 = 20 \]
This shows that the center of the circle is at \((4, 2)\) and the radius is \(\sqrt{20}\).

### Step 2: Find the equation of the tangent to the ellipse
The tangent to the ellipse at a point \((x_1, y_1)\) can be expressed as:
\[ \frac{x x_1}{16} + \frac{y y_1}{4} = 1 \]
Let’s denote the point on the ellipse as \((4 \cos \theta, 2 \sin \theta)\). Therefore, the equation of the tangent at this point becomes:
\[ \frac{x (4 \cos \theta)}{16} + \frac{y (2 \sin \theta)}{4} = 1 \]
Simplifying this gives:
\[ \cos \theta \cdot x + \sin \theta \cdot y = 1 \]

### Step 3: Check if the tangent is normal to the circle
For the tangent to be normal to the circle, it must pass through the center of the circle \((4, 2)\). Thus, substituting \(x = 4\) and \(y = 2\) into the tangent equation:
\[ \cos \theta \cdot 4 + \sin \theta \cdot 2 = 1 \]
This simplifies to:
\[ 4 \cos \theta + 2 \sin \theta = 1 \]

### Step 4: Solve for \(\theta\)
Rearranging the equation gives:
\[ 4 \cos \theta + 2 \sin \theta - 1 = 0 \]
Dividing through by 2:
\[ 2 \cos \theta + \sin \theta - \frac{1}{2} = 0 \]
This can be rewritten as:
\[ \sin \theta = -2 \cos \theta + \frac{1}{2} \]

### Step 5: Find possible values of \(\theta\)
To find the values of \(\theta\), we can use the identity:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Substituting \(\sin \theta\) into this identity will allow us to solve for \(\theta\). However, we can also analyze the equation \(2 \cos \theta + \sin \theta = \frac{1}{2}\) directly.

The equation \( \cos \theta + \sin \theta = 1 \) has solutions:
- \(\theta = 0\)
- \(\theta = \frac{\pi}{2}\)

### Final Answer
Thus, the values of \(\theta\) are:
\[ \theta = 0 \quad \text{or} \quad \theta = \frac{\pi}{2} \]