The area of rectangle formed by perpendiculars from the centre of ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1` to the tangent and normal at the point whose eccentric angle is `pi//4` is

Options

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

Answer

To find the area of the rectangle formed by the perpendiculars from the center of the ellipse to the tangent and normal at the point whose eccentric angle is \(\frac{\pi}{4}\), we can follow these steps:

### Step 1: Identify the coordinates of the point on the ellipse
The coordinates of a point on the ellipse corresponding to the eccentric angle \(\theta\) are given by:
\[ (x, y) = (a \cos \theta, b \sin \theta) \]
For \(\theta = \frac{\pi}{4}\):
\[ x = a \cos\left(\frac{\pi}{4}\right) = a \cdot \frac{1}{\sqrt{2}} = \frac{a}{\sqrt{2}} \]
\[ y = b \sin\left(\frac{\pi}{4}\right) = b \cdot \frac{1}{\sqrt{2}} = \frac{b}{\sqrt{2}} \]
So the coordinates of the point are \(\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\).

### Step 2: Find the equation of the tangent line
The equation of the tangent line to the ellipse at the point \((x_0, y_0)\) is given by:
\[ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1 \]
Substituting \(x_0 = \frac{a}{\sqrt{2}}\) and \(y_0 = \frac{b}{\sqrt{2}}\):
\[ \frac{x \cdot \frac{a}{\sqrt{2}}}{a^2} + \frac{y \cdot \frac{b}{\sqrt{2}}}{b^2} = 1 \]
Simplifying this gives:
\[ \frac{x}{\sqrt{2}a} + \frac{y}{\sqrt{2}b} = 1 \]
or
\[ x + \frac{a}{b}y = \sqrt{2}a \]

### Step 3: Find the equation of the normal line
The slope of the tangent line at the point is given by:
\[ -\frac{b^2 x_0}{a^2 y_0} \]
Substituting \(x_0\) and \(y_0\):
\[ -\frac{b^2 \cdot \frac{a}{\sqrt{2}}}{a^2 \cdot \frac{b}{\sqrt{2}}} = -\frac{b}{a} \]
The slope of the normal line is the negative reciprocal:
\[ \frac{a}{b} \]
Using the point-slope form of the line, the equation of the normal line is:
\[ y - \frac{b}{\sqrt{2}} = \frac{a}{b}\left(x - \frac{a}{\sqrt{2}}\right) \]

### Step 4: Find the intercepts of the tangent and normal lines
**Tangent line intercepts:**
- Setting \(y = 0\) in the tangent line equation:
\[ x + 0 = \sqrt{2}a \implies x = \sqrt{2}a \]
- Setting \(x = 0\) in the tangent line equation:
\[ 0 + \frac{a}{b}y = \sqrt{2}a \implies y = \sqrt{2}b \]

**Normal line intercepts:**
- Setting \(y = 0\) in the normal line equation:
\[ 0 - \frac{b}{\sqrt{2}} = \frac{a}{b}(x - \frac{a}{\sqrt{2}}) \implies x = \frac{b^2}{a} + \frac{a}{\sqrt{2}} \]
- Setting \(x = 0\) in the normal line equation:
\[ y - \frac{b}{\sqrt{2}} = \frac{a}{b}\left(0 - \frac{a}{\sqrt{2}}\right) \implies y = \frac{b}{\sqrt{2}} - \frac{a^2}{b\sqrt{2}} \]

### Step 5: Calculate the area of the rectangle
The area \(A\) of the rectangle formed by the intercepts is given by:
\[ A = \text{(length of tangent intercept)} \times \text{(length of normal intercept)} \]
Using the intercepts calculated in the previous step, we can find the lengths and thus the area.