Problem 184 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is [2006] (a) \(\frac{3}{5}\) (b) \(\frac{1}{2}\) (c) \(\frac{4}{5}\) (d) \(\frac{1}{\sqrt{5}}\)

Step-by-Step Solution

Answer

The eccentricity is \(\frac{3}{5}\), which is option (a).

Step 1: Understand properties of an ellipse

An ellipse with its center at the origin has the general equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. The distance between the foci is \(2c\), where \(c = \sqrt{a^2 - b^2}\). The eccentricity \(e\) is given by \(e = \frac{c}{a}\).

Step 2: Calculate the semi-minor axis, b

We are given that the minor axis is 8, so \(b = \frac{8}{2} = 4\).

Step 3: Calculate distance between foci, 2c

We know the distance between the foci is 6, so \(2c = 6\). This implies that \(c = \frac{6}{2} = 3\).

Step 4: Calculate value of a using c and b

Using the formula \(c = \sqrt{a^2 - b^2}\), substitute \(c = 3\) and \(b = 4\): \(3 = \sqrt{a^2 - 4^2}\).

Step 5: Solve for a

Square both sides to get \(9 = a^2 - 16\). Then solve for \(a^2\) by adding 16 to both sides: \(a^2 = 25\). Calculate \(a\): \(a = \sqrt{25} = 5\).

Step 6: Calculate eccentricity e

The eccentricity \(e\) is calculated as \(e = \frac{c}{a} = \frac{3}{5}\).

Step 7: Validate answer with given options

The answer \(e = \frac{3}{5}\) corresponds to option (a).

Key Concepts

Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being a perfect circle. A circle has an eccentricity of 0, while an ellipse has a value that is greater than 0 but less than 1.
Eccentricity is denoted by the letter \( e \) and can be calculated using the formula:

• \( e = \frac{c}{a} \)

where \( c \) is the distance from the center to a focus, and \( a \) is the semi-major axis of the ellipse.
A smaller eccentricity means the ellipse is more circular, while a larger eccentricity suggests it is more elongated. In our problem, where we found \( e = \frac{3}{5} \), the ellipse is moderately elongated.

Semi-Major Axis

The semi-major axis is the longest radius of an ellipse, extending from the center to the farthest edge. It is the distance denoted by \( a \). When thinking about an ellipse, imagine stretching a circle in one direction.
This stretch in the longer direction is described by the semi-major axis.
The formula that relates the semi-major axis to other parts of the ellipse is:

• \( c = \sqrt{a^2 - b^2} \)

where \( b \) is the semi-minor axis.
In our given exercise, solving for \( a \) using the equation \( 3 = \sqrt{a^2 - 4^2} \) provides us with \( a = 5 \). This means the semi-major axis measures 5 units.

Semi-Minor Axis

The semi-minor axis is the shortest radius of the ellipse, stretching from its center to the nearest edge. This measurement is denoted by \( b \). It is crucial in defining the shape of the ellipse.
If the semi-minor axis is much shorter than the semi-major axis, the ellipse will appear more stretched. The relationship among the semi-major axis, semi-minor axis, and the foci can be expressed as:

• \( b = \sqrt{a^2 - c^2} \)

The given problem mentions that the minor axis is 8. Therefore, the semi-minor axis \( b \) is \( \frac{8}{2} = 4 \). This means our ellipse is somewhat elongated.

Foci

The foci of an ellipse are two special points located along its major axis that play a key role in defining its shape. Any point on the ellipse has the same total distance to these two foci.
This is a defining property of ellipses. In formulas, when we say the distance between the foci is \( 2c \), \( c \) represents the distance from the center to one focus.
In our exercise, it is given that the distance between the foci is 6, hence \( 2c = 6 \), leading to \( c = 3 \).
These foci are vital in determining the eccentricity and hence the overall form of the ellipse.