if the tangents on the ellipse `4x^(2)+y^(2)=8` at the points (1,2) and (a,b) are perpendicular to each other then `a^(2)` is equal to

Options

A. `(128)/(17)`
B. `(64)/(17)`
C. `(4)/(17)`
D. `(2)/(17)`

Answer

To solve the problem, we need to find the value of \( a^2 \) given that the tangents to the ellipse \( 4x^2 + y^2 = 8 \) at the points \( (1, 2) \) and \( (a, b) \) are perpendicular to each other.

### Step-by-Step Solution:

1. **Find the equation of the ellipse:**
The given ellipse is \( 4x^2 + y^2 = 8 \).

2. **Differentiate the ellipse equation:**
To find the slope of the tangent line at any point on the ellipse, we differentiate implicitly:
\[ \frac{d}{dx}(4x^2 + y^2) = \frac{d}{dx}(8) \]
This gives:
\[ 8x + 2y \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 2y \frac{dy}{dx} = -8x \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{4x}{y} \]

3. **Find the slope at the point \( (1, 2) \):**
Substitute \( x = 1 \) and \( y = 2 \):
\[ m_1 = -\frac{4(1)}{2} = -2 \]

4. **Find the slope at the point \( (a, b) \):**
The slope at point \( (a, b) \) is:
\[ m_2 = -\frac{4a}{b} \]

5. **Set up the condition for perpendicular tangents:**
Since the tangents are perpendicular, we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting the values:
\[ (-2) \left(-\frac{4a}{b}\right) = -1 \]
Simplifying gives:
\[ \frac{8a}{b} = -1 \quad \Rightarrow \quad 8a = -b \quad \Rightarrow \quad b = -8a \]

6. **Substitute \( b \) into the ellipse equation:**
Since \( (a, b) \) lies on the ellipse, we substitute \( b = -8a \) into the ellipse equation:
\[ 4a^2 + (-8a)^2 = 8 \]
This simplifies to:
\[ 4a^2 + 64a^2 = 8 \quad \Rightarrow \quad 68a^2 = 8 \]

7. **Solve for \( a^2 \):**
Dividing both sides by 68:
\[ a^2 = \frac{8}{68} = \frac{2}{17} \]

### Final Answer:
Thus, the value of \( a^2 \) is \( \frac{2}{17} \).