Problem 104 - A Complete Resource Book in Mathematics for JEE Main

Question

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\), is: (A) 18 (B) \(\frac{27}{2}\) (C) 27 (D) \(\frac{27}{4}\)

Step-by-Step Solution

Answer

The area is 27 square units.

Step 1: Identify the Latus Rectum

The latus rectum of an ellipse is a line segment perpendicular to the major axis and passes through a focus. For the given ellipse \(\frac{x^2}{9} + \frac{y^2}{5} = 1\), the length of the latus rectum is \(\frac{2b^2}{a}\), where \(a^2 = 9\) and \(b^2 = 5\). In this case, the length of the latus rectum is \(\frac{2 \times 5}{3} = \frac{10}{3}\).

Step 2: Find Endpoints of the Latus Rectum

The endpoints of the latus rectum are points on the ellipse and can be found using the coordinates of the foci. The foci are located at \( (\pm c, 0)\), where \(c = \sqrt{a^2 - b^2} = \sqrt{9-5} = 2\). The endpoints of the latus rectum are \((\pm 2, \pm \frac{10}{3})\).

Step 3: Equation of Tangent at Endpoints

The equation of the tangent to an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at point \((x_1, y_1)\) is \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\). The tangents at the endpoints \((2, \frac{10}{3})\) and \((2, -\frac{10}{3})\) are: 1. \(\frac{x \times 2}{9} + \frac{y \times \frac{10}{3}}{5} = 1\) simplifies to \(\frac{2x}{9} + \frac{y}{3} = 1\).2. \(\frac{x \times 2}{9} - \frac{y \times \frac{10}{3}}{5} = 1\) simplifies to \(\frac{2x}{9} - \frac{y}{3} = 1\).

Step 4: Solve for the Intersection Points

To find the intersection points of the tangents, solve the equations \(\frac{2x}{9} + \frac{y}{3} = 1\) and \(\frac{2x}{9} - \frac{y}{3} = 1\) as a system. Adding these two equations results in \(\frac{4x}{9} = 2\), giving \(x = \frac{9}{2}\). Substituting \(x\) back, \(\frac{2(\frac{9}{2})}{9} + \frac{y}{3} = 1\) yields \(y = 0\). Thus, the intersection point is \(\left(\frac{9}{2}, 0\right)\). Subtracting gives another point, leading us to solve by symmetry, ultimately getting the vertices \((0, 3)\), \((0, -3)\), \((\frac{9}{2}, 0)\),, and \((-\frac{9}{2}, 0)\).

Step 5: Compute the Area of the Quadrilateral

Use the coordinates of the vertices obtained to calculate the area of the quadrilateral using the shoelace formula. These vertices will form a rectangle centered at the origin, with length \(9\) and width \(6\). Hence, the area is \(9 \times 6 = 54\). Divide by 2 (because it is half of a rectangle), leading to an area of \(27\).

Step 6: Conclusion

The area of the quadrilateral formed by the tangents is \(27\) square units.

Key Concepts

Latus Rectum

The latus rectum is a key concept in understanding ellipses, serving as a line segment that offers insights into the ellipse's dimensions. Situated perpendicular to the major axis and intersecting at a focus, the latus rectum is crucial for identifying specific points on the ellipse. For our given ellipse \( \frac{x^2}{9} + \frac{y^2}{5} = 1 \), the formula to determine the latus rectum's length is \( \frac{2b^2}{a} \). Here, \( a^2 = 9 \) and \( b^2 = 5 \), resulting in a length of \( \frac{10}{3} \). The touchpoints of the latus rectum are located at the foci's coordinates \((\pm 2, 0)\) alongside the vertical distance given by the latus rectum's length, so the coordinates are \((\pm 2, \pm \frac{10}{3})\).

Area of Quadrilateral

Understanding the area of a quadrilateral formed by tangents to an ellipse involves more than simple arithmetic. We found vertices such as \((0, 3)\), \((0, -3)\), \((\frac{9}{2}, 0)\), and \(( -\frac{9}{2}, 0)\), which represent crucial points where the tangents intersect. An easy method to calculate the area enclosed by these points is the shoelace formula, an efficient technique for determining areas given several vertices.
The rectangle formed at the origin, with its edges touching these vertices, has a length of 9 and a width of 6. The calculated area is \(9 \times 6 = 54\), but dividing by 2 corrects this as we are seeking the area of a quadrilateral—a result of 27 square units. This step by step thought process ensures the answer is both logical and accurate.

• Length of quad: 9
• Width of quad: 6
• Formula: \(\text{Area} = \frac{9 \times 6}{2} \)
• Final Area: 27 square units

Tangent to Ellipse

Tangents play an intriguing role in ellipse geometry. They serve as straight lines that touch the ellipse at a single point, without crossing it. Using tangent properties, we can derive equations to find the points of intersection. For the ellipse such as \(\frac{x^2}{9} + \frac{y^2}{5} = 1\), the tangent at any point \((x_1, y_1)\) is given by \( \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \).
We apply this concept at latus rectum's endpoints which yields two equations: \( \frac{2x}{9} + \frac{y}{3} = 1 \) and \( \frac{2x}{9} - \frac{y}{3} = 1 \). By solving these equations simultaneously, we find that they intersect at crucial points \((\frac{9}{2}, 0)\) and symmetrically at \((-\frac{9}{2}, 0)\) along with \((0, 3)\) and \((0, -3)\), which are the vertices of our quadrilateral. This understanding of how tangents operate aids in dissecting geometric problems involving ellipses.

• Equation Formulation: \( \frac{xx_1}{9} + \frac{yy_1}{5} = 1 \)
• Intersection points: \( (\frac{9}{2}, 0) \) and symmetry-based solutions
• Vertices Formed: Intersection vertices outline necessary tangential intersections