If a tangent of slope 4 of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is normal to the circle `x^(2)+y^(2)+4x+1=0` then the maximum value of `ab` is

Options

A. 4
B. 5
C. 6
D. 8

Explanation

To solve the problem, we need to find the maximum value of \( ab \) given the conditions of the ellipse and the circle.

### Step-by-step Solution:

1. **Identify the given equations**:
- The equation of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
- The equation of the circle is \(x^2 + y^2 + 4x + 1 = 0\).

2. **Rewrite the circle's equation**:
- We can rewrite the circle's equation in standard form:
\[ x^2 + 4x + y^2 + 1 = 0 \implies (x + 2)^2 + y^2 = 3 \]
- This shows that the center of the circle is \((-2, 0)\) and the radius is \(\sqrt{3}\).

3. **Find the equation of the tangent to the ellipse**:
- The slope of the tangent is given as \(m = 4\).
- The equation of the tangent to the ellipse at slope \(m\) is:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2 \]
- Substituting \(m = 4\):
\[ y = 4x \pm \sqrt{16a^2 + b^2} \]

4. **Normal to the circle**:
- The normal to the circle at any point passes through the center of the circle \((-2, 0)\).
- For the tangent line \(y = 4x + c\) to be normal to the circle, it must pass through the center \((-2, 0)\):
\[ 0 = 4(-2) + c \implies c = 8 \]
- Thus, the equation of the tangent becomes:
\[ y = 4x + 8 \quad \text{or} \quad y = 4x - 8 \]

5. **Setting the tangent equal to the ellipse**:
- For the tangent \(y = 4x + 8\) to touch the ellipse, we substitute \(y\) into the ellipse equation:
\[ \frac{x^2}{a^2} + \frac{(4x + 8)^2}{b^2} = 1 \]
- Expanding this gives:
\[ \frac{x^2}{a^2} + \frac{16x^2 + 64x + 64}{b^2} = 1 \]
- Rearranging leads to:
\[ \left(\frac{1}{a^2} + \frac{16}{b^2}\right)x^2 + \frac{64}{b^2}x + \left(\frac{64}{b^2} - 1\right) = 0 \]

6. **Condition for tangency**:
- For the above quadratic in \(x\) to have exactly one solution (tangency), the discriminant must be zero:
\[ \left(\frac{64}{b^2}\right)^2 - 4\left(\frac{1}{a^2} + \frac{16}{b^2}\right)\left(\frac{64}{b^2} - 1\right) = 0 \]

7. **Solving for \(ab\)**:
- After simplifying and solving the above equation, we find that:
\[ 16a^2 + b^2 = 64 \]
- Using the AM-GM inequality:
\[ \frac{16a^2 + b^2}{2} \geq \sqrt{16a^2 \cdot b^2} \]
- Thus:
\[ 32 \geq 4ab \implies ab \leq 8 \]

8. **Conclusion**:
- The maximum value of \(ab\) is \(8\).

### Final Answer:
The maximum value of \(ab\) is \(8\).