Chapter 11: Problem 181
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) \(3 x^{2}+5 y^{2}-32=0\)
Short Answer
Expert verified
The equation is (d) \( 3x^2 + 5y^2 - 32 = 0 \).
Step by step solution
01
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.
02
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, \[ \frac{b^2}{a^2} = \frac{3}{5}. \]
03
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 \[ 3x^2 + 5y^2 = 15k. \]
04
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 \( 3x^2 + 5y^2 = 32 \).
05
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).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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 \( k \).
- \( a \) 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.
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.