Problem 171 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\), is: (a) \(\frac{27}{2}\) (b) 27 (c) \(\frac{27}{4}\) (d) 18

Step-by-Step Solution

Answer

The area of the quadrilateral is 18 square units (d).

Step 1: Understanding the Ellipse Equation

The equation of the ellipse is \( \frac{x^2}{9} + \frac{y^2}{5} = 1 \). This is a standard form of the ellipse equation, where \( a^2 = 9 \) and \( b^2 = 5 \). Thus, \( a = 3 \) and \( b = \sqrt{5} \). This implies it's an ellipse with the semi-major axis along the x-axis and semi-minor axis along the y-axis.

Step 2: Identify the Length of Latus Rectum

The formula for the length of the latus rectum of an ellipse is given by \( \frac{2b^2}{a} \). Substituting the values of \( b^2 \) and \( a \), we get \( \frac{2 \times 5}{3} = \frac{10}{3} \).

Step 3: Equation of Tangents at End Points of Latus Rectum

The endpoints of the latus rectum for the ellipse are located at \((a, \pm \frac{b^2}{a}) = (3, \pm \frac{5}{3})\). The equation of a tangent to the ellipse at any point \((x_1, y_1)\) is \( \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \). For the points \((3, \frac{5}{3})\) and \((3, -\frac{5}{3})\), the tangents are \( \frac{3x}{9} + \frac{\frac{5}{3}y}{5} = 1 \) and \( \frac{3x}{9} - \frac{\frac{5}{3}y}{5} = 1 \).

Step 4: Simplify the Tangent Equations

Simplifying the tangent equations gives \( \frac{x}{3} + \frac{y}{3} = 1 \) and \( \frac{x}{3} - \frac{y}{3} = 1 \), or equivalently, \( x + y = 3 \) and \( x - y = 3 \).

Step 5: Find Intersection Points of Tangents

Using the equations from Step 4, the intersection of the lines \( x + y = 3 \) and \( x - y = 3 \). Solving these simultaneously, we get \( x = 3 \) and \( y = 0 \), \( x = 0 \) and \( y = 3 \). These will serve as diagonal points of the quadrilateral.

Step 6: Area Calculation of the Quadrilateral

Given that the quadrilateral formed by the tangents is a rectangle or symmetrical figure, we can find the area as half the product of its diagonals. Calculating the length of the diagonals using points \( (3, 0) \) and \((-3, 0) \), and \( (0, 3) \) and \( (0, -3) \), the diagonal lengths both measure 6. Therefore, the area is \( \frac{1}{2} \times 6 \times 6 = 18 \).

Key Concepts

Understanding the Latus Rectum

The latus rectum of an ellipse is a special line segment that plays a significant role in understanding the properties of the ellipse. It is parallel to the minor axis and crosses through one of the foci. The length of the latus rectum can be found using the formula \( \frac{2b^2}{a} \), where \(a\) and \(b\) are the semi-major and semi-minor axis of the ellipse, respectively.
For the ellipse with equation \( \frac{x^2}{9} + \frac{y^2}{5} = 1 \), \(a = 3\) and \(b = \sqrt{5}\). This gives the latus rectum length as:

• Substitute into the formula: \(\frac{2\times 5}{3} = \frac{10}{3} \).

This line helps establish key points for finding tangents, making it essential for deeper analysis of the ellipse.

Tangent to the Ellipse

Tangent lines to an ellipse are lines that touch the ellipse at one single point without crossing it. They have unique properties because they form a right angle with the radius at the point of contact. The general equation for the tangent at a point \((x_1, y_1)\) on an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is:

• \( \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \).

For the ellipse in this example, we find the tangent lines at the endpoints of the latus rectum:

• The points are \((3, \frac{5}{3})\) and \((3, -\frac{5}{3})\).
• The tangents are \(x + y = 3\) and \(x - y = 3\), respectively.

These lines are essential in understanding how segments can enclose spaces like quadrilaterals.

Calculating the Area of a Quadrilateral

When finding the area of a quadrilateral formed by tangents to an ellipse, it's valuable to determine the shape's vertices or diagonal points. The key here is solving for the intersection of tangent lines. From our previous calculations:

• The tangent lines \(x + y = 3\) and \(x - y = 3\) intersect at points \((3, 0)\) and \((0, 3)\), making them two opposing vertices.

To find the area, we assume symmetry. The quadrilateral resembles a rectangle, enabling a straightforward area calculation by measuring diagonals:

• Each diagonal length is 6, as derived from points like \((3, 0)\) and \((-3, 0)\).
• Formula for area: \(\frac{1}{2} \times \text{Diagonal}_1 \times \text{Diagonal}_2 = 18\).

Understanding this process helps ease the analysis of any similar geometric problem involving ellipses.