To solve the problem, we need to analyze each statement one by one and determine their validity.
### Statement 1:
**Statement:** The locus of the centroid of a triangle formed by three co-normal points on a parabola is the axis of the parabola.
**Solution Steps:**
1. **Understanding Co-normal Points:** Co-normal points on a parabola are points where the normals to the parabola intersect at a common point.
2. **Equation of the Parabola:** Let’s consider the parabola \( y^2 = 4ax \).
3. **Normal to the Parabola:** The equation of the normal at a point \( (at^2, 2at) \) on the parabola is given by:
\[
y - 2at = -\frac{1}{t}(x - at^2)
\]
Rearranging gives us:
\[
y = -\frac{1}{t}x + 2a + \frac{at^2}{t}
\]
4. **Finding the Centroid:** For three co-normal points \( A, B, C \) with parameters \( t_1, t_2, t_3 \):
- The coordinates of these points are:
\[
A(at_1^2, 2at_1), \quad B(at_2^2, 2at_2), \quad C(at_3^2, 2at_3)
\]
- The centroid \( G \) of triangle \( ABC \) is given by:
\[
G\left(\frac{at_1^2 + at_2^2 + at_3^2}{3}, \frac{2a(t_1 + t_2 + t_3)}{3}\right)
\]
5. **Locus of the Centroid:** As \( t_1, t_2, t_3 \) vary, the x-coordinate of the centroid depends on the squares of the parameters, while the y-coordinate is linear in the parameters. However, the sum of the parameters \( t_1 + t_2 + t_3 \) can be made to approach zero, leading to the conclusion that the locus of the centroid lies on the x-axis, which is the axis of the parabola.
**Conclusion:** Statement 1 is **True**.
### Statement 2:
**Statement:** One of the angles between the parabolas \( y^2 = 8x \) and \( x^2 = 27y \) is \( \tan^{-1}\left(\frac{9}{13}\right) \).
**Solution Steps:**
1. **Finding the Slopes:**
- For \( y^2 = 8x \), we differentiate to find \( \frac{dy}{dx} \):
\[
2y \frac{dy}{dx} = 8 \implies \frac{dy}{dx} = \frac{4}{y}
\]
- For \( x^2 = 27y \), we differentiate:
\[
2x = 27 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2x}{27}
\]
2. **Finding Points of Intersection:** Set \( y^2 = 8x \) and substitute into \( x^2 = 27y \) to find intersection points.
3. **Calculating the Angles:** Use the slopes \( m_1 = \frac{4}{y} \) and \( m_2 = \frac{2x}{27} \) at the points of intersection to find the angle \( \theta \):
\[
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
\]
4. **Verification:** After substituting the values and simplifying, we find that one of the angles indeed equals \( \tan^{-1}\left(\frac{9}{13}\right) \).
**Conclusion:** Statement 2 is **True**.
### Statement 3:
**Statement:** The product of lengths of perpendiculars drawn from foci to a tangent of the ellipse \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) is 4.
**Solution Steps:**
1. **Identify the Foci:** For the ellipse \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \):
- Semi-major axis \( a = 3 \), semi-minor axis \( b = 2 \).
- The foci are located at \( (\pm c, 0) \) where \( c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5} \).
2. **Equation of Tangent:** The equation of the tangent at point \( (x_0, y_0) \) on the ellipse is:
\[
\frac{xx_0}{9} + \frac{yy_0}{4} = 1
\]
3. **Length of Perpendiculars:** The length of the perpendicular from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
4. **Calculating Product:** Calculate the lengths of the perpendiculars from both foci to the tangent line and find their product. After simplification, we find that the product equals 4.
**Conclusion:** Statement 3 is **True**.
### Final Conclusion:
All three statements are **True**.
