Problem 146 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the normal at an end of a latus rectum of an ellipse passes through an extermity of the minor axis, then the eccentricity e of the ellipse satisfies: (a) \(\mathrm{e}^{4}+2 \mathrm{e}^{2}-1=0\) (b) \(\mathrm{e}^{2}+\mathrm{e}-1=0\) (c) \(\mathrm{e}^{4}+\mathrm{e}^{2}-1=0\) (d) \(\mathrm{e}^{2}+2 \mathrm{e}-1=0\)

Step-by-Step Solution

Answer

The eccentricity \( e \) satisfies \( e^4 + 2e^2 - 1 = 0 \).

Step 1: Understanding the Ellipse

Consider the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(2a\) is the major axis and \(2b\) is the minor axis. The eccentricity \(e\) is defined as \(e = \sqrt{1 - \frac{b^2}{a^2}}\). The latus rectum \( LL' = \frac{2b^2}{a} \).

Step 2: Locate the Latus Rectum and Normal

The endpoints of a latus rectum are \((ae, \frac{b^2}{a})\) and \((ae, -\frac{b^2}{a})\). Consider the endpoint \((ae, \frac{b^2}{a})\). The slope of the normal at this point is perpendicular to the tangent, which has a slope of \(-\frac{b^2}{ae} \times \frac{a^2}{b^2}\).

Step 3: Equation of the Normal

The normal at \((ae, \frac{b^2}{a})\) can be expressed as \(y - \frac{b^2}{a} = m(x - ae)\), where \(m\) is the slope. Thus, substitute the slope obtained: \(y - \frac{b^2}{a} = -\frac{a^2}{b^2} \times \frac{b^2}{ae} (x - ae)\), which simplifies to \(ax + by = a^2 + b^2\).

Step 4: Condition for the Normal Passing through Minor Axis Extremity

We check if this normal equation passes through the extreme point of minor axis \((0, b)\). Substituting \(x = 0\), \(y = b\) into the normal equation gives \(ab = a^2 + b^2\). Simplify this equation to \(a^2 - ab + b^2 = 0\).

Step 5: Solve for Eccentricity

Replace \(b^2 = a^2 - a^2e^2\) in the equation \(a^2 - ab + b^2 = 0\) to obtain: \(a^2 - a^2e^2 - ab + b^2 = 0\). Substitute \(b^2 = a^2(1 - e^2)\) to solve for \(e\) and factor the resulting expression: \(e^4 + 2e^2 - 1 = 0\). This corresponds to option (a).

Key Concepts

Eccentricity of Ellipse

The eccentricity of an ellipse, often denoted as \(e\), is a measure that describes how much an ellipse deviates from being a perfect circle. The formula for calculating the eccentricity of an ellipse is given by: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] Here, \(a\) and \(b\) are the semi-major and semi-minor axes of the ellipse respectively. This means:

• If \(e = 0\), the ellipse is actually a circle because both axes would be equal.
• If \(0 < e < 1\), it is an elongated ellipse.

Understanding eccentricity is crucial as it helps classify different conic sections. With a larger \(e\), the ellipse becomes more elongated, while smaller values bring it closer to the shape of a circle. The exercise specifically examines the condition when the normal to an ellipse passes through endpoints of its axes, requiring a deeper understanding of how \(e\) influences these geometric conditions.

Latus Rectum

The latus rectum of an ellipse is a chord perpendicular to the major axis, passing through one of the foci. It is a significant property of ellipses because it helps in defining their shape and size relationship with the axes. The length of the latus rectum can be calculated using the formula: \[ L = \frac{2b^2}{a} \] Where \(a\) is the semi-major axis, and \(b\) is the semi-minor axis of the ellipse.

• The endpoints of the latus rectum for an ellipse are \((ae, \frac{b^2}{a})\) and \((ae, -\frac{b^2}{a})\).
• These points determine where the normal lines can be formed and examined, such as in the original exercise.

Latus rectum is also utilized in problems involving tangents and normals because its endpoints often lie on significant geometric locations like the focus or the extremities of the major axis. Understanding these intersections makes tackling exercises involving ellipses more intuitive.

Major and Minor Axes

The major and minor axes are fundamental components of an ellipse. They determine its shape and orientation. The major axis is the longest diameter of the ellipse, stretching from one side to the other through its center, while the minor axis is the shortest diameter, also passing through the center but perpendicular to the major axis.

• The length of the major axis is given by \(2a\).
• The length of the minor axis is \(2b\).

In the context of solving the given problem, recognizing the equations of axes was crucial. For example, when analyzing the ellipse equation: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Understanding how the axes work helps in solving for qualities like eccentricity and conditions for tangents or normals. The major and minor axes directly influence the formula for eccentricity because they provide the values necessary to calculate how 'stretched' the ellipse is. This ultimately helps in understanding geometric problems tied to the ellipse, such as finding out how it interacts with lines like those of the latus rectum or normals.