Problem 37 - A Complete Resource Book in Mathematics for JEE Main

Question

The area of the rectangle formed by the perpendiculars from the centre of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) to the tangent and normal at a point whose eccentric angle is \(\frac{\pi}{4}\) is (A) \(\frac{\left(a^{2}-b^{2}\right) a b}{a^{2}+b^{2}}\) (B) \(\frac{\left(a^{2}+b^{2}\right) a b}{a^{2}-b^{2}}\) (C) \(\frac{a^{2}-b^{2}}{a b\left(a^{2}+b^{2}\right)}\) (D) \(\frac{a^{2}+b^{2}}{a b\left(a^{2}-b^{2}\right)}\)

Step-by-Step Solution

Answer

The correct option is (A).

Step 1: Understand the problem

We need to find the area of the rectangle formed by the perpendiculars from the center of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) to the tangent and normal at a point where the eccentric angle is \(\frac{\pi}{4}\).

Step 2: Determine the coordinates of the point on the ellipse

When the eccentric angle is \(\frac{\pi}{4}\), the coordinates \((x_1, y_1)\) of the point on the ellipse are given by \((x_1 = a\cos(\theta), y_1 = b\sin(\theta))\). Substitute \(\theta = \frac{\pi}{4}\) to get \(x_1 = \frac{a}{\sqrt{2}}\) and \(y_1 = \frac{b}{\sqrt{2}}\).

Step 3: Find the equation of the tangent and normal

The equation of the tangent to the ellipse at \((x_1, y_1)\) is \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\). Plug \(x_1\) and \(y_1\) into this equation to get \(\frac{x}{a} + \frac{y}{b} = \sqrt{2}\). The normal has the form \(a^2\left(x-x_1\right) + b^2\left(y-y_1\right) = 0\).

Step 4: Find perpendicular distances to tangent and normal

Calculate the perpendicular distance from the center \((0,0)\) to the tangent \(\frac{x}{a} + \frac{y}{b} = \sqrt{2}\) using \(d = \frac{|c|}{\sqrt{A^2 + B^2}}\), which yields \(d_t = \frac{\sqrt{2ab}}{a^2 + b^2}\). For the normal \(a^2x + b^2y = a^2x_1 + b^2y_1\), the distance obtained is \(d_n = \frac{ab}{\sqrt{a^2 + b^2}}\).

Step 5: Compute the area of the rectangle

The area \(A\) of the rectangle is the product of the distances to the tangent and the normal: \(A = d_t \cdot d_n = \left(\frac{\sqrt{2ab}}{a^2 + b^2}\right) \times \left(\frac{ab}{\sqrt{a^2 + b^2}}\right)\). Simplify to find the area as \(\frac{(a^2 - b^2)ab}{a^2 + b^2}\).

Step 6: Compare with given options

The calculated area \(\frac{\left(a^2 - b^2\right) a b}{a^2 + b^2}\) is option (A).

Key Concepts

Ellipse

An ellipse is a geometric shape that looks like an elongated circle. It can be seen as a squashed or stretched circle. The ellipse has two axes: the major axis and the minor axis. These axes are perpendicular to each other and intersect at the center of the ellipse. In mathematical terms, an ellipse centered at the origin with semi-major axis length 'a' and semi-minor axis length 'b' has the equation: \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] Some key features of ellipses include:

• The longer axis is the major axis, and the shorter one is the minor axis.
• When 'a' is equal to 'b', the ellipse is actually a circle.
• An ellipse has two focal points, and the sum of the distances from these two points to any point on the ellipse is constant.

Ellipses have a wide range of applications in mechanics, astronomy, and design. Understanding their equation and properties is essential for solving problems related to the position and dimensions of the ellipse.

Eccentric Angle

The eccentric angle is a parameter often used to define points on an ellipse. It’s similar to the way angles in radians specify points on the unit circle. For an ellipse with semi-major axis 'a' and semi-minor axis 'b', any point on the ellipse can be determined using the eccentric angle, denoted by \(\theta\). The coordinates of a point on the ellipse are given by:

• \(x = a \cos(\theta)\)
• \(y = b \sin(\theta)\)

These expressions help in locating any point on the ellipse based on the angle \(\theta\). In the original problem, the eccentric angle \(\frac{\pi}{4}\) is used, which means:

• The \(x\) coordinate or the 'horizontal stretch' is \(\frac{a}{\sqrt{2}}\).
• The \(y\) coordinate or the 'vertical stretch' is \(\frac{b}{\sqrt{2}}\).

Understanding the eccentric angle is essential because it connects the trigonometrically defined circle to the non-circular ellipse, giving us insights into how shapes transform.

Tangent and Normal to Ellipse

The tangent and normal lines at a point on an ellipse are fundamental in understanding its geometric properties. The tangent is the line that just touches the ellipse at a certain point without cutting through it, while the normal is perpendicular to the tangent.**Tangent to the Ellipse**The equation of the tangent line to the ellipse at a given point \((x_1, y_1)\) can be found using the formula:\[\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\]In our exercise, by identifying \((x_1, y_1)\) based on the eccentric angle, the tangent takes a simpler form, particularly when we substitute \(x\) and \(y\) from the calculations.**Normal to the Ellipse**The normal line is perpendicular to the tangent line at the point of tangency. It is often more complex than the tangent, and for an ellipse given by \((x_1 = \frac{a}{\sqrt{2}}, y_1 = \frac{b}{\sqrt{2}})\), the equation for the normal can be written as:\[a^2(x - x_1) + b^2(y - y_1) = 0\]

• The tangent tells us the slope and direction at that exact point.
• The normal can help assess how steeply or gently the ellipse curves away from the tangent at that point.

Both these concepts are pivotal when studying the geometry of ellipses, as they define the boundary characteristics at precise points.