Problem 154 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The length of the minor axis (along \(y\)-axis) of an ellipse in the standard form is \(\frac{4}{\sqrt{3}}\). If this ellipse touches the line, \(x\) \(+6 y=8\); then its eccentricity is: (a) \(\frac{1}{2} \sqrt{\frac{11}{3}}\) (b) \(\sqrt{\frac{5}{6}}\) (c) \(\frac{1}{2} \sqrt{\frac{5}{3}}\) (d) \(\frac{1}{3} \sqrt{\frac{11}{3}}\)

Step-by-Step Solution

Answer

The eccentricity is option (a): \(\frac{1}{2} \sqrt{\frac{11}{3}}\).

Step 1: Equation of the ellipse

The standard form of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Given, the length of the minor axis is \(\frac{4}{\sqrt{3}}\), hence \(2b \frac{4}{\sqrt{3}} = 2b\), implying \(b = \frac{2}{\sqrt{3}}\).

Step 2: Calculate the distance from the line to the center

The distance \(d\) from the center of the ellipse, which is at \((0,0)\), to the line \(x + 6y = 8\) is calculated using the formula \(\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\). Substituting\(A = 1\), \(B = 6\), \(C = -8\), \(x_1 = 0\), \(y_1 = 0\), we get \(d = \frac{|0 + 0 - 8|}{\sqrt{1^2 + 6^2}} = \frac{8}{\sqrt{37}}\).

Step 3: Relate distance to semi-major axis

Since the ellipse touches the line, the distance \(d\) equals the semi-major axis \(a\). Thus, \(a = \frac{8}{\sqrt{37}}\).

Step 4: Calculate eccentricity

The eccentricity \(e\) is given by \(e = \frac{\sqrt{a^2 - b^2}}{a}\). We found \(b = \frac{2}{\sqrt{3}}\) and \(a = \frac{8}{\sqrt{37}}\). Calculate \(a^2 = \left(\frac{8}{\sqrt{37}}\right)^2 = \frac{64}{37}\) and \(b^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}\). Hence, \(e = \sqrt{\frac{64}{37} - \frac{4}{3}} = \sqrt{\frac{64}{37} - \frac{49.333}{37}} = \sqrt{\frac{14.667}{37}} = \frac{1}{2} \sqrt{\frac{11}{3}}\).

Step 5: Final Calculation and Comparison

Finally, the calculated eccentricity is \(\frac{1}{2} \sqrt{\frac{11}{3}}\), which corresponds to option (a).

Key Concepts

Eccentricity

Ellipses have an interesting property known as eccentricity, denoted by \(e\), which measures how much the shape deviates from being a circle. Eccentricity helps describe the shape of the ellipse.
For an ellipse, the eccentricity is between 0 and 1.
A circle has an eccentricity of 0, while an ellipse which stretches to a more elongated shape towards becoming a parabola has an eccentricity closer to 1. The formula for finding the eccentricity of an ellipse is given by:

• \(e = \frac{\sqrt{a^2 - b^2}}{a}\)
• Where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

In the given exercise, knowing the values of \(a^2\) and \(b^2\), you can compute the eccentricity. Each term contributes valuable insights into how the shape of the ellipse varies as eccentricity changes.

Minor Axis

The minor axis of an ellipse is the shortest diameter that runs through its center, while the longest diameter is the major axis.
In mathematical terms, the minor axis is associated with the smaller denominator in the standard form of an ellipse's equation: \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
Here, \(b\) is the length of the semi-minor axis. However, the entire minor axis has a total length of \(2b\).
In this exercise, the minor axis is specified along the y-axis with a total length \(\frac{4}{\sqrt{3}}\), leading to:

• \(2b = \frac{4}{\sqrt{3}}\)
• Solving, \(b = \frac{2}{\sqrt{3}}\)

The minor axis often helps determine the proportion and physical spread of an ellipse, playing a crucial role in its geometry.

Major Axis

The major axis of an ellipse is its longest diameter passing through its center, making it critical in defining an ellipse's size.
In the standard ellipse equation, the term \(a\) relates to the semi-major axis length:

• \(a^2 > b^2\) always holds since the major axis is longer.

This axis plays a key role in determining eccentricity and overall proportions.
In our scenario, the major axis is identical to the line's distance from the ellipse's center since the ellipse touches the line.
The distance is calculated using a specific formula:\[d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\]
With values, \(a = \frac{8}{\sqrt{37}}\), representing the semi-major axis where it touches the line. A major axis extends to a full length of \(2a\), contributing significantly to the shape and characteristics of the ellipse.

Touching a Line

When an ellipse touches a line, it means the line acts as a tangent to the ellipse, making contact at exactly one point. This geometric property is crucial in various fields, from design to physics.
For the exercise in question, the ellipse touches the line represented by the equation \(x + 6y = 8\).
The significant consequence of touching a line is that the perpendicular distance from the ellipse's center to the line equals the semi-major axis \(a\).
The formula used to calculate the perpendicular distance from a point (in this case, the center) to a line is:

• \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
• Where \((x_1, y_1)\) is the ellipse center, and \(A\), \(B\), and \(C\) correlate to the line coefficients \(Ax + By + C = 0\).

When the ellipse touches a line, it means that the equation's overlap equates to one contact point, signifying a tangent. This geometric interaction further helps solve equations involving ellipses, ensuring elegant and exact solutions.