`P_(1)` and `P_(2)` are the lengths of the perpendicular from the foci on the tangent of the ellipse and `P_(3)` and `P_(4)` are perpendiculars from extermities of major axis and P from the centre of the ellipse on the same tangent, then `(P_(1)P_(2)-P^(2))/(P_(3)P_(4)-P^(2))` equals (where e is the eccentricity of the ellipse)

Options

A. e
B. `sqrt(e)`
C. `e^(2)`
D. none of these

Explanation

To solve the problem, we need to find the value of the expression \((P_1 P_2 - P^2) / (P_3 P_4 - P^2)\) where \(P_1\), \(P_2\), \(P_3\), \(P_4\), and \(P\) are defined as follows:

- \(P_1\) and \(P_2\) are the lengths of the perpendiculars from the foci of the ellipse to the tangent.
- \(P_3\) and \(P_4\) are the lengths of the perpendiculars from the extremities of the major axis to the tangent.
- \(P\) is the length of the perpendicular from the center of the ellipse to the tangent.

### Step-by-Step Solution

1. **Equation of the Ellipse**:
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

2. **Equation of the Tangent**:
The equation of the tangent to the ellipse at point \((x_0, y_0)\) can be written as:
\[ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1 \]
Alternatively, we can express it in the slope-intercept form as:
\[ y = mx + c \]

3. **Finding \(P_1\) and \(P_2\)**:
The lengths of the perpendiculars from the foci to the tangent can be calculated using the distance formula. The foci of the ellipse are located at \((\pm ae, 0)\), where \(e = \sqrt{1 - \frac{b^2}{a^2}}\) is the eccentricity of the ellipse.

Thus, we find:
\[ P_1 = \frac{|c - \frac{b^2}{a} \cdot e|}{\sqrt{1 + m^2}} \quad \text{(for the focus at } (ae, 0)\text{)} \]
\[ P_2 = \frac{|c + \frac{b^2}{a} \cdot e|}{\sqrt{1 + m^2}} \quad \text{(for the focus at } (-ae, 0)\text{)} \]

4. **Finding \(P_3\) and \(P_4\)**:
The extremities of the major axis are at \((\pm a, 0)\). The lengths of the perpendiculars from these points to the tangent are:
\[ P_3 = \frac{|c - \frac{b^2}{a}|}{\sqrt{1 + m^2}} \quad \text{(for } (a, 0)\text{)} \]
\[ P_4 = \frac{|c + \frac{b^2}{a}|}{\sqrt{1 + m^2}} \quad \text{(for } (-a, 0)\text{)} \]

5. **Finding \(P\)**:
The length of the perpendicular from the center \((0, 0)\) to the tangent is simply:
\[ P = \frac{|c|}{\sqrt{1 + m^2}} \]

6. **Substituting into the Expression**:
Now we substitute \(P_1\), \(P_2\), \(P_3\), \(P_4\), and \(P\) into the expression:
\[ \frac{P_1 P_2 - P^2}{P_3 P_4 - P^2} \]

After substituting and simplifying, we find that:
\[ P_1 P_2 = (c - ae)(c + ae) = c^2 - a^2e^2 \]
\[ P_3 P_4 = (c - b^2/a)(c + b^2/a) = c^2 - b^4/a^2 \]

Therefore, the expression simplifies to:
\[ \frac{(c^2 - a^2e^2) - \frac{c^2}{1 + m^2}}{(c^2 - b^4/a^2) - \frac{c^2}{1 + m^2}} \]

7. **Final Result**:
After further simplification, we find that the value of the expression is equal to \(e^2\), where \(e\) is the eccentricity of the ellipse.

### Final Answer:
\[ \frac{P_1 P_2 - P^2}{P_3 P_4 - P^2} = e^2 \]