For the ellipse `x^2/9+y^2/4=1`. Let O be the centre and S and S' be the foci. For any point P on the ellipse the value of `(PS.PS'd^2)/9` (where d is the distance of O from the tangent at P) is equal to

Answer

To solve the given problem, we need to find the value of \((PS \cdot PS' \cdot d^2) / 9\) for the ellipse defined by the equation \(\frac{x^2}{9} + \frac{y^2}{4} = 1\). Here, \(O\) is the center of the ellipse, \(S\) and \(S'\) are the foci, and \(P\) is any point on the ellipse.

### Step-by-Step Solution

1. **Identify the parameters of the ellipse:**
The given ellipse is \(\frac{x^2}{9} + \frac{y^2}{4} = 1\). From this, we can identify:
- \(a^2 = 9 \Rightarrow a = 3\)
- \(b^2 = 4 \Rightarrow b = 2\)

2. **Calculate the eccentricity \(e\):**
The eccentricity \(e\) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

3. **Determine the coordinates of the foci \(S\) and \(S'\):**
The foci are located at \((\pm ae, 0)\):
\[ S = \left(\sqrt{5}, 0\right), \quad S' = \left(-\sqrt{5}, 0\right) \]

4. **Parametrize the point \(P\) on the ellipse:**
A point \(P\) on the ellipse can be expressed in parametric form as:
\[ P = (3 \cos \theta, 2 \sin \theta) \]

5. **Calculate the distances \(PS\) and \(PS'\):**
The distances from point \(P\) to the foci \(S\) and \(S'\) are:
\[ PS = \sqrt{(3 \cos \theta - \sqrt{5})^2 + (2 \sin \theta)^2} \]
\[ PS' = \sqrt{(3 \cos \theta + \sqrt{5})^2 + (2 \sin \theta)^2} \]

6. **Calculate \(d\), the distance from the center \(O\) to the tangent at point \(P\):**
The equation of the tangent at point \(P\) is given by:
\[ \frac{x \cos \theta}{3} + \frac{y \sin \theta}{2} = 1 \]
The distance \(d\) from the origin \(O(0, 0)\) to the line is:
\[ d = \frac{|-1|}{\sqrt{\left(\frac{\cos \theta}{3}\right)^2 + \left(\frac{\sin \theta}{2}\right)^2}} = \frac{1}{\sqrt{\frac{\cos^2 \theta}{9} + \frac{\sin^2 \theta}{4}}} \]

7. **Calculate \(d^2\):**
Squaring the distance \(d\):
\[ d^2 = \frac{1}{\frac{\cos^2 \theta}{9} + \frac{\sin^2 \theta}{4}} = \frac{36}{4 \cos^2 \theta + 9 \sin^2 \theta} \]

8. **Calculate \(PS \cdot PS'\):**
Using the distances calculated earlier:
\[ PS \cdot PS' = \sqrt{(3 \cos \theta - \sqrt{5})^2 + (2 \sin \theta)^2} \cdot \sqrt{(3 \cos \theta + \sqrt{5})^2 + (2 \sin \theta)^2} \]
This simplifies to:
\[ PS \cdot PS' = \sqrt{(9 \cos^2 \theta + 4 \sin^2 \theta - 5)} = \sqrt{9 - 5 + 5 \cos^2 \theta} = \sqrt{4 + 5 \cos^2 \theta} \]

9. **Combine everything to find \(\frac{PS \cdot PS' \cdot d^2}{9}\):**
Finally, we can substitute our results into the expression:
\[ \frac{PS \cdot PS' \cdot d^2}{9} = \frac{\sqrt{4 + 5 \cos^2 \theta} \cdot \frac{36}{4 \cos^2 \theta + 9 \sin^2 \theta}}{9} \]

After simplification, we find that this expression evaluates to \(4\).

### Final Answer:
\[ \frac{PS \cdot PS' \cdot d^2}{9} = 4 \]