Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

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

(A) \(\frac{(a^2 - b^2) ab}{a^2 + b^2}\)

Step 1: Determine the Parametric Coordinates

For the given ellipse \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), the parametric coordinates can be written as \(x = a \cos \theta\) and \(y = b \sin \theta\). Here, the eccentric angle \(\theta\) is given as \(\frac{\pi}{4}\). Plugging in, we get \(x = \frac{a}{\sqrt{2}}\) and \(y = \frac{b}{\sqrt{2}}\).

Step 2: Find the Slope of the Tangent and Normal

The slope of the tangent line at any point on the ellipse is given by \(-\frac{b^{2}}{a^{2}} \cdot \frac{x}{y}\). Substituting the parametric coordinates, we get the slope of the tangent as \(-\frac{b^{2}}{a^{2}} \cdot \frac{a}{b} = -\frac{b}{a}\). The slope of the normal is the negative reciprocal: \(\frac{a}{b}\).

Step 3: Write the Equations of Tangent and Normal

The equation of the tangent at the point with coordinates \((\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}})\) is \(bx - ay + \frac{(a^{2} - b^{2})}{\sqrt{2}} = 0\). Similarly, the equation for the normal is \(ax + by - ab = 0\).

Step 4: Calculate the Perpendicular Distance from the Center

The perpendicular distance from the center (0,0) to the tangent is \(\frac{|\frac{(a^{2} - b^{2})}{\sqrt{2}}|}{\sqrt{a^{2} + b^{2}}}\). To the normal, it is \(\frac{|ab|}{\sqrt{a^{2} + b^{2}}}\).

Step 5: Evaluate the Area of the Rectangle

The area of the rectangle is the product of the two perpendicular distances calculated in step 4. Therefore, it is: \[ \frac{(a^{2} - b^{2})}{\sqrt{2}\sqrt{a^{2} + b^{2}}} \times \frac{ab}{\sqrt{a^{2} + b^{2}}} = \frac{(a^{2} - b^{2})ab}{a^{2} + b^{2}}.\]

Step 6: Choose the Correct Answer

The correct answer matches option (A): \(\frac{(a^{2} - b^{2}) ab}{a^{2} + b^{2}}\).

Key Concepts

Parametric Equations

Understanding parametric equations can significantly simplify working with ellipses. An ellipse is traditionally expressed as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), which can be complex to deal with directly in certain scenarios. Parametric equations convert this into simpler forms using trigonometric functions:

• For the ellipse, the parametric forms are \( x = a \cos \theta \) and \( y = b \sin \theta \).
• The parameter \( \theta \) is known as the eccentric angle, representing specific points on the ellipse as it changes.

In the given exercise, we have the eccentric angle \( \theta = \frac{\pi}{4} \), allowing us to easily find the coordinates \( x = \frac{a}{\sqrt{2}} \) and \( y = \frac{b}{\sqrt{2}} \). This approach replaces the variables with manageable trigonometric terms.

Tangent and Normal Slopes

Finding the slopes of tangent and normal lines to an ellipse is crucial for understanding its geometry. The tangent line touches the ellipse at a single point, while the normal line is perpendicular to the tangent.

• First, the slope of the tangent line can be calculated using the formula: \( -\frac{b^2}{a^2} \times \frac{x}{y} \).
• By substituting the parametric coordinates, we determine: \( -\frac{b^2}{a^2} \times \frac{a}{b} = -\frac{b}{a} \).
• The normal slope, which is the negative reciprocal of the tangent slope, will be \( \frac{a}{b} \).

Having these slopes helps us derive the equations for the tangent and normal lines. This understanding is fundamental to solving further geometric problems, such as finding perpendicular distances or calculating areas.

Perpendicular Distances

Once the equations of the tangent and normal lines are determined, it's valuable to know how to compute their perpendicular distances from the ellipse's center. This distance measurement reveals how far a point or object is from the line in a perpendicular direction.

• For the tangent line \( bx - ay + \frac{(a^{2} - b^{2})}{\sqrt{2}} = 0 \), the perpendicular distance from the center (0,0) is: \( \frac{|\frac{(a^2 - b^2)}{\sqrt{2}}|}{\sqrt{a^2 + b^2}} \).
• Similarly, for the normal line \( ax + by - ab = 0 \), the perpendicular distance is: \( \frac{|ab|}{\sqrt{a^2 + b^2}} \).

These calculations are handy, especially when setting up problems involving areas or lengths, where the measurement along the perpendicular axis is necessary.

Rectangle Area Calculation

Calculating the area of the rectangle formed by these distances involves understanding basic geometry principles.

• The area of a rectangle is straightforward: multiply its length by its width.
• Here, the length and width are the perpendicular distances from the center to the tangent and normal lines, calculated as: \( \frac{(a^2 - b^2)}{\sqrt{2}\sqrt{a^2 + b^2}} \) and \( \frac{ab}{\sqrt{a^2 + b^2}} \) respectively.
• Multiplying these two distances gives the rectangle's area as: \[ \frac{(a^2 - b^2)ab}{a^2 + b^2} \]

This formula reflects the interactions of the ellipse’s geometric properties, helping students to derive answers to more complex geometry problems involving ellipses.