The sum of the squares of the perpendiculars on any tangent to the ellipse a 2 x 2 ​ + b 2 y 2 ​ =1 from two points on the minor axis, each at a distances ae from the centre, is

Options

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

Answer

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 to any tangent of the ellipse. The ellipse is given by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 1: Identify the Points on the Minor Axis

The minor axis of the ellipse is along the y-axis. The points given are at a distance \( ae \) from the center, which means the points are:

\[ (0, ae) \quad \text{and} \quad (0, -ae) \]

### Step 2: Write the Equation of the Tangent

The equation of any tangent to the ellipse can be expressed as:

\[ y = mx + \sqrt{a^2 m^2 + b^2} \]

where \( m \) is the slope of the tangent.

### Step 3: Calculate the Perpendicular Distance from Each Point to the Tangent

The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the tangent line \( y = mx + \sqrt{a^2 m^2 + b^2} \), we can rearrange it to the standard form:

\[ mx - y + \sqrt{a^2 m^2 + b^2} = 0 \]

Here, \( A = m \), \( B = -1 \), and \( C = \sqrt{a^2 m^2 + b^2} \).

### Step 4: Calculate the Distance from the Points

1. For the point \( (0, ae) \):

\[ d_1 = \frac{|m(0) - ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} = \frac{|-ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} \]

2. For the point \( (0, -ae) \):

\[ d_2 = \frac{|m(0) - (-ae) + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} = \frac{|ae + \sqrt{a^2 m^2 + b^2}|}{\sqrt{m^2 + 1}} \]

### Step 5: Calculate the Sum of the Squares of the Distances

Now, we need to find the sum of the squares of these distances:

\[ d_1^2 + d_2^2 = \left(\frac{-ae + \sqrt{a^2 m^2 + b^2}}{\sqrt{m^2 + 1}}\right)^2 + \left(\frac{ae + \sqrt{a^2 m^2 + b^2}}{\sqrt{m^2 + 1}}\right)^2 \]

This simplifies to:

\[ \frac{(-ae + \sqrt{a^2 m^2 + b^2})^2 + (ae + \sqrt{a^2 m^2 + b^2})^2}{m^2 + 1} \]

### Step 6: Expand and Simplify

Expanding both squares:

\[ (-ae + \sqrt{a^2 m^2 + b^2})^2 = a^2 e^2 - 2ae\sqrt{a^2 m^2 + b^2} + (a^2 m^2 + b^2) \]

\[ (ae + \sqrt{a^2 m^2 + b^2})^2 = a^2 e^2 + 2ae\sqrt{a^2 m^2 + b^2} + (a^2 m^2 + b^2) \]

Adding these gives:

\[ 2a^2 e^2 + 2(a^2 m^2 + b^2) \]

Thus, we have:

\[ d_1^2 + d_2^2 = \frac{2(a^2 e^2 + a^2 m^2 + b^2)}{m^2 + 1} \]

### Step 7: Substitute \( e^2 \)

Recall that \( e^2 = 1 - \frac{b^2}{a^2} \), so we substitute \( e^2 \) into our equation:

\[ d_1^2 + d_2^2 = \frac{2\left(a^2(1 - \frac{b^2}{a^2}) + a^2 m^2 + b^2\right)}{m^2 + 1} \]

This simplifies to:

\[ d_1^2 + d_2^2 = \frac{2\left(a^2 - b^2 + a^2 m^2 + b^2\right)}{m^2 + 1} = \frac{2a^2(1 + m^2)}{m^2 + 1} \]

### Final Result

This results in:

\[ d_1^2 + d_2^2 = 2a^2 \]

Thus, the sum of the squares of the perpendiculars on any tangent to the ellipse from the two points on the minor axis is:

\[ \boxed{2a^2} \]