Problem 181 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point \((-3,1)\) and has eccentricity \(\sqrt{\frac{2}{5}}\) is [2011] (a) \(5 x^{2}+3 y^{2}-48=0\) (b) \(3 x^{2}+5 y^{2}-15=0\) (c) \(5 x^{2}+3 y^{2}-32=0\) (d) \(3 x^{2}+5 y^{2}-32=0\)

Step-by-Step Solution

Answer

The equation is (d) \( 3x^2 + 5y^2 - 32 = 0 \).

Step 1: Understanding the General Equation of an Ellipse

The standard form of an ellipse equation aligned with the coordinate axes is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively. We need to determine which type of ellipse it is based on the given conditions.

Step 2: Determine the Eccentricity Relation

Eccentricity \( e \) of an ellipse is given by \( e = \sqrt{1 - \frac{b^2}{a^2}} \). We are given that \( e = \sqrt{\frac{2}{5}} \). Thus, \[ \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{\frac{2}{5}}. \] Squaring both sides, \[ 1 - \frac{b^2}{a^2} = \frac{2}{5}. \] Simplifying, \[ \frac{b^2}{a^2} = \frac{3}{5}. \]

Step 3: Set Up the Equation

From \( \frac{b^2}{a^2} = \frac{3}{5} \), we can let \( b^2 = 3k \) and \( a^2 = 5k \) for some \( k \). Hence, the equation becomes \( \frac{x^2}{5k} + \frac{y^2}{3k} = 1 \), or equivalently, \( \frac{3x^2}{15k} + \frac{5y^2}{15k} = 1 \). Multiplying through by \( 15k \), we get \[ 3x^2 + 5y^2 = 15k. \]

Step 4: Check Which Equation Satisfies the Point (-3,1)

Substitute \( x = -3 \) and \( y = 1 \) into the equation \( 3x^2 + 5y^2 = 15k \):\[ 3(-3)^2 + 5(1)^2 = 27 + 5 = 32. \] Hence, \( 15k = 32 \). The equation of the ellipse is \( 3x^2 + 5y^2 = 32 \).

Step 5: Match with Provided Options

Review the options given in the problem:(a) \( 5x^2 + 3y^2 - 48 = 0 \)(b) \( 3x^2 + 5y^2 - 15 = 0 \)(c) \( 5x^2 + 3y^2 - 32 = 0 \)(d) \( 3x^2 + 5y^2 - 32 = 0 \)The derived equation \( 3x^2 + 5y^2 = 32 \) corresponds to option (d).

Key Concepts

Eccentricity of Ellipse

The eccentricity of an ellipse is a measure that describes how much the ellipse deviates from being a perfect circle. An ellipse's eccentricity, denoted as \( e \), is defined as \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] where \( a \) is the length of the semi-major axis and \( b \) is the length of the semi-minor axis.
Ellipses can have eccentricity values between 0 and 1:

• When \( e = 0 \), the ellipse is actually a circle. This is because both axes are equal in length.
• As \( e \) approaches 1, the ellipse becomes more elongated.

In our exercise, we are given \( e = \sqrt{\frac{2}{5}} \). This helps us find the relationship between \( a \) and \( b \) through the equation \[ \frac{b^2}{a^2} = \frac{3}{5} \]. This fraction tells us that the semi-minor axis is shorter compared to the semi-major axis, indicating an elongated ellipse rather than a circular shape.

Standard Form of Ellipse

The standard form of the equation of an ellipse with axes parallel to the coordinate axes plays a crucial role in identifying and working with ellipses. For an ellipse centered at the origin, the standard equation is given as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here:

• \( a \) is the semi-major axis, which is the longest radius of the ellipse.
• \( b \) is the semi-minor axis, the shortest radius of the ellipse.

These axes are always perpendicular to each other.
In our task, we understood that \( \frac{b^2}{a^2} = \frac{3}{5} \). By expressing \( b^2 = 3k \) and \( a^2 = 5k \), we maintained this ratio and arrived at \( \frac{x^2}{5k} + \frac{y^2}{3k} = 1 \). Multiplying by \( 15k \), we deduced the form \( 3x^2 + 5y^2 = 15k \), which simplifies the equation of the ellipse in the problem after substituting the known point and solving for \( k \).

Coordinate Geometry of Ellipses

Coordinate geometry provides a straightforward approach to visualize and solve problems involving ellipses using a coordinate plane. With ellipses, we frequently encounter the standard form equations aligned with coordinate axes, assisting in identifying the ellipse's orientation and dimensions.

• The general setup involves the ellipse centered around the origin, making equations symmetric concerning the x and y-axes.
• It allows for easily determining distances such as semi-major and semi-minor axes from a single focal point (origin) and efficiently calculating the eccentricity.

By using coordinate geometry principles, we better understand distances and relationships involving points on the ellipse. For the given problem, once aligned in this geometric framework, checking a point like \((-3, 1)\) fits seamlessly to validate which given equation correctly models the ellipse, leading us to derive and confirm the correct option.