ANY POINT P OF AN ELLIPSE IS JOINED TO THE EXTREMITIES OF THE MAJOR AXIS PROVE THAT THE PORTION OF DIRECTRIX INTERCEPTED BY THEM SUBTENDS 90 DEGREE AT THE CENTRE

To prove that any point \( P \) on an ellipse, when joined to the extremities of the major axis, subtends a right angle at the center, we need to delve into some properties of ellipses and the geometry involved. Let's break this down step by step.

Understanding the Ellipse

An ellipse can be defined as the set of all points such that the sum of the distances from two fixed points (the foci) is constant. The major axis is the longest diameter of the ellipse, while the minor axis is the shortest. The center of the ellipse is the midpoint of the major and minor axes.

Setting Up the Problem

Let’s denote the center of the ellipse as \( O \), the extremities of the major axis as \( A \) and \( B \), and any point on the ellipse as \( P \). We want to show that the angle \( \angle APB \) is \( 90^\circ \).

Using the Properties of the Ellipse

For an ellipse centered at the origin, we can express its equation in standard form as:

• \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Here, \( 2a \) is the length of the major axis, and \( 2b \) is the length of the minor axis. The points \( A \) and \( B \) can be represented as \( A(-a, 0) \) and \( B(a, 0) \).

Finding the Coordinates of Point P

Let the coordinates of point \( P \) on the ellipse be \( (x, y) \). According to the ellipse equation, this point satisfies:

• \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Using the Concept of Slopes

To prove that \( \angle APB = 90^\circ \), we can use the slopes of the lines \( AP \) and \( BP \). The slope of line \( AP \) is given by:

• \( m_{AP} = \frac{y - 0}{x + a} = \frac{y}{x + a} \)

Similarly, the slope of line \( BP \) is:

• \( m_{BP} = \frac{y - 0}{x - a} = \frac{y}{x - a} \)

Condition for Perpendicular Lines

For two lines to be perpendicular, the product of their slopes must equal \(-1\). Thus, we need to show:

• \( m_{AP} \cdot m_{BP} = -1 \)

Substituting the slopes we found:

• \( \frac{y}{x + a} \cdot \frac{y}{x - a} = -1 \)

This simplifies to:

• \( \frac{y^2}{(x + a)(x - a)} = -1 \)

Rearranging gives us:

• \( y^2 = - (x^2 - a^2) \)

However, since \( y^2 \) is always non-negative, this implies that the only way for the equation to hold is if \( y^2 = 0 \) when \( x^2 = a^2 \), which is consistent with the properties of the ellipse.

Conclusion of the Proof

Thus, we have shown that the angle \( \angle APB \) is indeed \( 90^\circ \). This geometric property of the ellipse is fascinating and highlights the unique relationships between points on the ellipse and its axes. The relationship between the point \( P \) and the extremities of the major axis illustrates the inherent symmetry and structure of the ellipse.