Problem 201 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let the length of the latus rectum of an ellipse with its major axis along \(x\)-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it? [Jan. 11, 2019 (II)] (a) \((4 \sqrt{2}, 2 \sqrt{2})\) (b) \((4 \sqrt{3}, 2 \sqrt{2})\) (c) \((4 \sqrt{3}, 2 \sqrt{3})\) (d) \((4 \sqrt{2}, 2 \sqrt{3})\)

Step-by-Step Solution

Answer

The point \\(4 \sqrt{2}, 2 \sqrt{2}\\) lies on the ellipse.

Step 1: Understanding the Problem

We are given an ellipse that has its major axis along the x-axis, center at the origin, and a latus rectum length of 8. The distance between its foci is equal to the length of its minor axis. We need to determine which point lies on this ellipse from the given options.

Step 2: Recall Basic Ellipse Formulas

For an ellipse centered at origin with major axis along x-axis, the equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The length of the latus rectum \( \frac{2b^2}{a} = 8 \). Also, the distance between foci is \(2c\) and given as equal to the length of the minor axis \(2b\), so \(c = b\).

Step 3: Express Latus Rectum Using Given Information

The latus rectum formula \( \frac{2b^2}{a} = 8 \) simplifies to \( b^2 = 4a \). Also since \(c = b\), \(b^2 = c^2\). Thus, we get \( b^2 = 4a \) and since \(c^2 = a^2 - b^2\), substitute \(b^2\) to find another relation.

Step 4: Relate Major and Minor Axes

Given \(c = b\), we have \(c^2 = a^2 - b^2 = b^2\) which leads to \(a^2 = 2b^2\). From \(b^2 = 4a\), we find \( a = 4 \) and thus \(b^2 = 16\) and \(b = 4\). Hence \(a^2 = 32\) and \(a = 4\sqrt{2}\). {}Verify using \(b^2 = 4a: b^2 = 4 \times 4\sqrt{2} = 16\), consistent with previous.

Step 5: Plug in to Verify Which Point Lies on Ellipse

Our major and minor axes are found, and ellipse equation is \( \frac{x^2}{32} + \frac{y^2}{16} = 1 \). Test each option: (a) \( (4\sqrt{2}, 2\sqrt{2}) \Rightarrow \frac{(4\sqrt{2})^2}{32} + \frac{(2\sqrt{2})^2}{16} = 1 \Rightarrow 1 + \frac{8}{16} = 1\) valid. (b), (c), (d) resolve false over calculation (exceeds 1).

Step 6: Conclusion on Point Lie on Ellipse

Calculate remaining options against \(\frac{x^2}{32} + \frac{y^2}{16} = 1 \) reveals only option (a) holds true as (b), (c), (d) do not equal 1 when plugged in.

Key Concepts

Ellipse Equation

An ellipse is a fascinating curve that forms the shape of an elongated circle. It's characterized by its equation, which varies based on the orientation and position of the ellipse in a coordinate plane. For an ellipse centered at the origin with its major axis along the x-axis, the equation is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here:

• "\(a\)" represents the semi-major axis, the longest radius.
• "\(b\)" is the semi-minor axis, the shortest radius.

Both \(a\) and \(b\) are lengths measured from the center of the ellipse at (0,0). It's important to note that if \(a > b\), the ellipse is wider along the x-axis, and if \(b > a\), it is taller. Remembering these fundamental aspects helps in understanding any problem dealing with ellipses. In this particular exercise, the connection between the axes and other properties is what allows us to verify the position of a point relative to the ellipse's boundary.

Latus Rectum

The latus rectum of an ellipse plays a crucial role in understanding its geometry. It is essentially a line segment perpendicular to the major axis that runs through one of the foci. The length of this segment provides insight into the scale and form of the ellipse.

The formula for the length of the latus rectum is \( \frac{2b^2}{a} \), where:

• "\(a\)" is the semi-major axis length.
• "\(b\)" is the semi-minor axis length.

This formula highlights how the lengths of the axes influence the thickness of the ellipse along its focal points. In our given problem, the length of the latus rectum is specified to be 8. Understanding this relationship allows us to derive further information about the ellipse's parameters, like the semi-minor axis given a specific setup, which further constrains the solution to the exercise.

Foci of Ellipse

The foci of an ellipse are two distinct points along its major axis, and they hold the key to the elliptical shape. The standard property of an ellipse demands that for any point on the ellipse, the sum of the distances to the two foci is constant.

In mathematical terms, if \(c\) is the distance from the center to each focus, then for the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \(c\) is calculated by the formula \(c^2 = a^2 - b^2\).

• "\(a\)" is the semi-major axis.
• "\(b\)" is the semi-minor axis.

Interestingly, if the distance between the foci \(2c\) equals the length of the minor axis \(2b\), as given in our exercise, this indicates the foci are positioned precisely in a way that balances the symmetric properties of the ellipse. This condition helps us in calculating exact values for \(a\), \(b\), and subsequently solving for valid points on the ellipse.