Ellipse
Equation of Ellipse
Grade None
Question:
<p>Equation of the ellipse whose axes are the axes of coordinates and which passes through the point \((-3, 1)\) and has eccentricity \(\sqrt{2/5}\) is</p>
<p>\(5x^2 + 3y^2 - 48 = 0\)</p>
<p>\(3x^2 + 5y^2 - 15 = 0\)</p>
<p>\(5x^2 + 3y^2 - 32 = 0\)</p>
<p>\(3x^2 + 5y^2 - 32 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the eccentricity condition e² = 1 - b²/a² along with the point condition to create a system of equations. For an ellipse with coordinate axes as axes, if a > b, then e² = 1 - b²/a².
<p><strong>Step 1:</strong> Let the ellipse equation be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b > 0$ (major axis along x-axis).</p><p><strong>Step 2:</strong> Use eccentricity condition: $e^2 = \frac{2}{5} = 1 - \frac{b^2}{a^2}$</p><p>This gives: $\frac{b^2}{a^2} = 1 - \frac{2}{5} = \frac{3}{5}$, so $b^2 = \frac{3a^2}{5}$</p><p><strong>Step 3:</strong> Since the ellipse passes through $(-3, 1)$, substitute into the equation:</p><p>$\frac{9}{a^2} + \frac{1}{b^2} = 1$</p><p><strong>Step 4:</strong> Substitute $b^2 = \frac{3a^2}{5}$:</p><p>$\frac{9}{a^2} + \frac{1}{\frac{3a^2}{5}} = 1$</p><p>$\frac{9}{a^2} + \frac{5}{3a^2} = 1$</p><p>$\frac{27 + 5}{3a^2} = 1$</p><p>$\frac{32}{3a^2} = 1$</p><p>$a^2 = \frac{32}{3}$ and $b^2 = \frac{3}{5} \cdot \frac{32}{3} = \frac{32}{5}$</p><p><strong>Step 5:</strong> The ellipse equation is: $\frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}} = 1$ or equivalently $\frac{3x^2}{32} + \frac{5y^2}{32} = 1$</p><p>Multiplying by 32: $3x^2 + 5y^2 = 32$</p><p>∴ Answer: D</p>
Correct Answer: D