Statement 1 Feet of prependiculars drawn from foci of an ellipse `4x^(2)+y^(2)=16` on the line `2sqrt3x+y=8` lie on the circle `x^(2)+y^(2)=16`
Statement 2 If prependiculars are from foci of an ellipse to its any tangent, the feet of these perpendicular lie on director circle of the ellipse.

Answer

To solve the problem, we need to analyze both statements regarding the ellipse given by the equation \(4x^2 + y^2 = 16\), the line \(2\sqrt{3}x + y = 8\), and the circle \(x^2 + y^2 = 16\).

### Step 1: Rewrite the equations in standard forms
1. **Ellipse**: The equation \(4x^2 + y^2 = 16\) can be rewritten as:
\[ \frac{x^2}{4} + \frac{y^2}{16} = 1 \]
This is an ellipse centered at the origin with semi-major axis 4 (along y-axis) and semi-minor axis 2 (along x-axis).

2. **Line**: The line \(2\sqrt{3}x + y = 8\) can be rewritten as:
\[ y = -2\sqrt{3}x + 8 \]
This is a straight line with a slope of \(-2\sqrt{3}\).

3. **Circle**: The equation \(x^2 + y^2 = 16\) represents a circle centered at the origin with a radius of 4.

### Step 2: Find the foci of the ellipse
The foci of the ellipse can be calculated using the formula:
\[ c = \sqrt{a^2 - b^2} \]
where \(a = 4\) and \(b = 2\). Thus:
\[ c = \sqrt{4^2 - 2^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \]
The foci are located at \((0, \pm 2\sqrt{3})\).

### Step 3: Find the feet of the perpendiculars from the foci to the line
To find the feet of the perpendiculars from the foci to the line \(2\sqrt{3}x + y = 8\), we can use the formula for the distance from a point to a line and the projection of the point onto the line.

1. **For the focus at \((0, 2\sqrt{3})\)**:
- The line can be expressed in the form \(Ax + By + C = 0\) as:
\[ 2\sqrt{3}x + y - 8 = 0 \]
- The distance \(D\) from the point \((0, 2\sqrt{3})\) to the line is given by:
\[ D = \frac{|2\sqrt{3}(0) + 1(2\sqrt{3}) - 8|}{\sqrt{(2\sqrt{3})^2 + 1^2}} = \frac{|2\sqrt{3} - 8|}{\sqrt{12 + 1}} = \frac{|2\sqrt{3} - 8|}{\sqrt{13}} \]
- To find the foot of the perpendicular, we can use the point-line distance formula and find the coordinates.

2. **For the focus at \((0, -2\sqrt{3})\)**:
- Similarly, we can calculate the foot of the perpendicular from this focus to the line.

### Step 4: Check if the feet lie on the circle
After finding the coordinates of the feet of the perpendiculars, we can check if these points satisfy the equation of the circle \(x^2 + y^2 = 16\).

### Conclusion
1. **Statement 1**: If the feet of the perpendiculars from the foci lie on the circle, then Statement 1 is **true**.
2. **Statement 2**: The statement regarding the feet of perpendiculars from the foci to any tangent lying on the director circle is **false** in this context.