From any point on the line `(t+2)(x+y) =1, t ne -2`, tangents are drawn to the ellipse `4x^(2)+16y^(2) = 1`. It is given that chord of contact passes through a fixed point. Then the number of integral values of 't' for which the fixed point always lies inside the ellipse is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem step by step, we will break down the given information and derive the necessary equations.

### Step 1: Understand the given line and ellipse
The line is given by the equation:
\[ (t + 2)(x + y) = 1 \quad \text{for } t \neq -2 \]
We can rewrite this as:
\[ x + y = \frac{1}{t + 2} \]

The ellipse is given by:
\[ 4x^2 + 16y^2 = 1 \]

### Step 2: Find the chord of contact
For a point \((\alpha, \beta)\) on the line, the chord of contact to the ellipse can be expressed as:
\[ 4x\alpha + 16y\beta = 1 \]
Substituting \(\beta = \frac{1}{t + 2} - \alpha\) into the chord of contact equation:
\[ 4x\alpha + 16y\left(\frac{1}{t + 2} - \alpha\right) = 1 \]
Expanding this gives:
\[ 4x\alpha + \frac{16y}{t + 2} - 16y\alpha = 1 \]
Rearranging terms:
\[ (4x - 16y)\alpha + \frac{16y}{t + 2} - 1 = 0 \]

### Step 3: Identify the fixed point
The chord of contact must pass through a fixed point. The coefficients of \(\alpha\) must equal zero for the line to pass through a fixed point:
\[ 4x - 16y = 0 \quad \Rightarrow \quad x = 4y \]
Substituting \(x = 4y\) into the equation gives:
\[ \frac{16y}{t + 2} - 1 = 0 \quad \Rightarrow \quad 16y = t + 2 \quad \Rightarrow \quad y = \frac{t + 2}{16} \]
Then substituting back to find \(x\):
\[ x = 4y = \frac{4(t + 2)}{16} = \frac{t + 2}{4} \]
Thus, the fixed point is:
\[ \left(\frac{t + 2}{4}, \frac{t + 2}{16}\right) \]

### Step 4: Substitute fixed point into the ellipse equation
We need this fixed point to lie inside the ellipse:
\[ 4\left(\frac{t + 2}{4}\right)^2 + 16\left(\frac{t + 2}{16}\right)^2 < 1 \]
Calculating this:
\[ 4 \cdot \frac{(t + 2)^2}{16} + 16 \cdot \frac{(t + 2)^2}{256} < 1 \]
Simplifying:
\[ \frac{(t + 2)^2}{4} + \frac{(t + 2)^2}{16} < 1 \]
Finding a common denominator:
\[ \frac{4(t + 2)^2 + (t + 2)^2}{16} < 1 \]
This simplifies to:
\[ \frac{5(t + 2)^2}{16} < 1 \]
Multiplying both sides by 16:
\[ 5(t + 2)^2 < 16 \]
Dividing by 5:
\[ (t + 2)^2 < \frac{16}{5} \]

### Step 5: Solve the inequality
Taking the square root:
\[ -\sqrt{\frac{16}{5}} < t + 2 < \sqrt{\frac{16}{5}} \]
This gives:
\[ -4/\sqrt{5} < t + 2 < 4/\sqrt{5} \]
Subtracting 2:
\[ -\frac{4}{\sqrt{5}} - 2 < t < \frac{4}{\sqrt{5}} - 2 \]

### Step 6: Calculate the bounds for \(t\)
Calculating the numerical bounds:
\[ -\frac{4}{\sqrt{5}} \approx -1.79 \quad \Rightarrow \quad -1.79 - 2 \approx -3.79 \]
\[ \frac{4}{\sqrt{5}} \approx 1.79 \quad \Rightarrow \quad 1.79 - 2 \approx -0.21 \]
Thus, we have:
\[ -3.79 < t < -0.21 \]

### Step 7: Find integral values of \(t\)
The integral values of \(t\) in this range are:
\[ -3, -2, -1 \]
Since \(t \neq -2\), the valid integral values are:
\[ -3, -1 \]

### Conclusion
Thus, the number of integral values of \(t\) for which the fixed point always lies inside the ellipse is:
\[ \boxed{2} \]