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:

• A. \( 5x^2 + 3y^2 - 48 = 0 \)
• B. \( 3x^2 + 5y^2 - 15 = 0 \)
• C. \( 5x^2 + 3y^2 - 32 = 0 \)
• D. \( 3x^2 + 5y^2 - 32 = 0 \)
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Hint:

To solve problems involving the equation of an ellipse, first determine the relationship between the semi-major and semi-minor axes using the given eccentricity, and then substitute the point given to find the value of \( a^2 \).

Answer 1

The Correct Option is D

The general equation of the ellipse whose axes are along the coordinate axes is given by: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \] where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively. The eccentricity \( e \) of the ellipse is related to \( a \) and \( b \) by the formula: \[ e = \sqrt{1 - \frac{b^2}{a^2}}. \] We are given that \( e = \sqrt{\frac{2}{5}} \), so: \[ \sqrt{\frac{2}{5}} = \sqrt{1 - \frac{b^2}{a^2}}, \] \[ \frac{2}{5} = 1 - \frac{b^2}{a^2}, \] \[ \frac{b^2}{a^2} = \frac{3}{5}. \] Thus, the relation between \( a^2 \) and \( b^2 \) is: \[ b^2 = \frac{3}{5}a^2. \] The equation of the ellipse now becomes: \[ \frac{x^2}{a^2} + \frac{y^2}{\frac{3}{5}a^2} = 1. \] Next, we substitute the point (-3, 1) into this equation to find the value of \( a^2 \). Substituting \( x = -3 \) and \( y = 1 \): \[ \frac{(-3)^2}{a^2} + \frac{1^2}{\frac{3}{5}a^2} = 1, \] \[ \frac{9}{a^2} + \frac{5}{3a^2} = 1, \] \[ \frac{27}{3a^2} + \frac{5}{3a^2} = 1, \] \[ \frac{32}{3a^2} = 1, \] \[ a^2 = \frac{32}{3}. \] Now, substitute \( a^2 = \frac{32}{3} \) into the equation of the ellipse: \[ \frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{3}{5} \times \frac{32}{3}} = 1, \] \[ \frac{3x^2}{32} + \frac{5y^2}{32} = 1. \] Multiply the entire equation by 32: \[ 3x^2 + 5y^2 = 32. \] Thus, the equation of the ellipse is: \[ 3x^2 + 5y^2 - 32 = 0. \].