The sum fo the squares of the perpendicular on any tangent to the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` from two points on the mirror axis, each at a distance `sqrt(a^(2) - b^(2))` from the centre, is

Options

A. `2a^(2)`
B. `2b^(2)`
C. `a^(2) + b^(2)`
D. `a^(2) - b^(2)`

Explanation

To solve the problem, we need to find the sum of the squares of the perpendiculars from two points on the minor axis of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) to any tangent line to the ellipse. The points are located at a distance of \( \sqrt{a^2 - b^2} \) from the center along the minor axis.

### Step-by-Step Solution:

1. **Identify the Points on the Minor Axis:**
The minor axis of the ellipse is along the y-axis. The points on the minor axis at a distance of \( \sqrt{a^2 - b^2} \) from the center (0,0) are:
\[ P(0, ae) \quad \text{and} \quad Q(0, -ae) \]
where \( e = \sqrt{1 - \frac{b^2}{a^2}} \) is the eccentricity of the ellipse.

2. **Equation of the Tangent to the Ellipse:**
The equation of any tangent to the ellipse can be expressed in the form:
\[ y = mx + \sqrt{a^2 m^2 + b^2} \]
Rearranging gives us:
\[ mx - y + \sqrt{a^2 m^2 + b^2} = 0 \]

3. **Calculate the Length of the Perpendicular from Point P:**
The length of the perpendicular \( p_1 \) from point \( P(0, ae) \) to the tangent line is given by the formula:
\[ p_1 = \frac{|mx_1 - y_1 + c|}{\sqrt{m^2 + 1}} \]
Substituting \( x_1 = 0 \) and \( y_1 = ae \):
\[ p_1 = \frac{|m(0) - ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} = \frac{|-ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} \]

4. **Calculate the Length of the Perpendicular from Point Q:**
Similarly, for point \( Q(0, -ae) \):
\[ p_2 = \frac{|m(0) - (-ae) + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} = \frac{|ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} \]

5. **Sum of the Squares of the Perpendiculars:**
Now we need to find \( p_1^2 + p_2^2 \):
\[ p_1^2 + p_2^2 = \left( \frac{|-ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} \right)^2 + \left( \frac{|ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} \right)^2 \]
This simplifies to:
\[ p_1^2 + p_2^2 = \frac{|-ae + \sqrt{a^2 m^2 + b^2}|^2 + |ae + \sqrt{a^2 m^2 + b^2}|^2}{m^2 + 1} \]

6. **Expanding the Squares:**
Let \( k = \sqrt{a^2 m^2 + b^2} \):
\[ p_1^2 + p_2^2 = \frac{(k - ae)^2 + (k + ae)^2}{m^2 + 1} \]
Expanding this gives:
\[ = \frac{(k^2 - 2k \cdot ae + a^2e^2) + (k^2 + 2k \cdot ae + a^2e^2)}{m^2 + 1} = \frac{2k^2 + 2a^2e^2}{m^2 + 1} \]

7. **Substituting \( k \):**
Recall \( k^2 = a^2 m^2 + b^2 \):
\[ p_1^2 + p_2^2 = \frac{2(a^2 m^2 + b^2) + 2a^2e^2}{m^2 + 1} \]

8. **Simplifying Further:**
Since \( b^2 = a^2(1 - e^2) \):
\[ p_1^2 + p_2^2 = \frac{2a^2(m^2 + b^2 + e^2)}{m^2 + 1} \]
Noting that \( b^2 + a^2e^2 = a^2 \):
\[ = \frac{2a^2}{m^2 + 1} \]

9. **Final Result:**
The final result simplifies to:
\[ p_1^2 + p_2^2 = 2a^2 \]

### Final Answer:
The sum of the squares of the perpendiculars from the two points on the minor axis to any tangent to the ellipse is \( 2a^2 \).