Understanding the Problem

We are to find the equation of an ellipse whose axes are aligned with the coordinate axes (i.e., major and minor axes are along x and y axes). It passes through the point (-3, 1) and has eccentricity $e=\sqrt{\frac{2}{5}}$.

Step 1: General Equation of the Ellipse

Since the axes are along the coordinate axes, the standard equation of the ellipse can be either:

Case 1: Major axis along x-axis: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a>b$ and eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$.

Case 2: Major axis along y-axis: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, where $a>b$ and eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$.

Given $e=\sqrt{\frac{2}{5}}$, which is less than 1, confirming it is an ellipse.

Step 2: Use the Eccentricity Condition

For both cases, we have $e^2=1-\frac{b^2}{a^2}$. Substituting $e=\sqrt{\frac{2}{5}}$:

$\frac{2}{5}=1-\frac{b^2}{a^2}$

Rearranging: $\frac{b^2}{a^2}=1-\frac{2}{5}=\frac{3}{5}$

So, $b^2=\frac{3}{5}{a}^2$.

Step 3: Use the Point (-3, 1) Condition

The ellipse passes through (-3, 1). We need to test both cases.

Case 1: Major axis along x-axis

Equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Substitute x = -3, y = 1 and $b^2=\frac{3}{5}{a}^2$:

$\frac{9}{a^2}+\frac{1}{\frac{3}{5}}=1$

Simplify: $\frac{9}{a^2}+\frac{5}{3}=1$

Multiply both sides by $a^2$: $9+\frac{5}{3}=a^2$

$\frac{27}{3}+\frac{5}{3}=a^2$ ⇒ $\frac{32}{3}=a^2$

Then $b^2=\frac{3}{5}\times \frac{32}{3}=\frac{32}{5}$

So the equation becomes: $\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1$

Multiply both sides by 32: $\frac{3}{32}{x}^2\times 32+\frac{5}{32}{y}^2\times 32=32$ ⇒ $3x^2+5y^2=32$

Or $3x^2+5y^2-32=0$

This matches one of the options.

Case 2: Major axis along y-axis

Equation: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Substitute x = -3, y = 1 and $b^2=\frac{3}{5}{a}^2$:

$\frac{9}{\frac{3}{5}}+\frac{1}{a^2}=1$

Simplify: $\frac{45}{3}+\frac{1}{a^2}=1$ ⇒ $\frac{15}{a^2}+\frac{1}{a^2}=1$ ⇒ $\frac{16}{a^2}=1$

So $a^2=16$, and $b^2=\frac{3}{5}\times 16=\frac{48}{5}$

The equation becomes: $\frac{x^2}{\frac{48}{5}}+\frac{y^2}{16}=1$

Multiply both sides by 48: $\frac{5}{48}{x}^2\times 48+\frac{1}{16}{y}^2\times 48=48$ ⇒ $5x^2+3y^2=48$

Or $5x^2+3y^2-48=0$

This also matches an option, but we must check which case is valid with the eccentricity condition. In Case 2, we assumed major axis along y-axis, so $a>b$. Here $a^2=16$ and $b^2=\frac{48}{5}=9.6$, so $a>b$ is satisfied. Both cases are mathematically possible, but we need to see which option is listed. The options are:

• 5x² + 5y² – 48 = 0
• 5x² + 3y² – 32 = 0
• 3x² + 5y² – 15 = 0
• 3x² + 5y² – 32 = 0

Our Case 1 gave 3x² + 5y² – 32 = 0, and Case 2 gave 5x² + 3y² – 48 = 0 (which is not exactly listed, but similar to 5x² + 3y² – 32 = 0). However, only 3x² + 5y² – 32 = 0 is an exact match from Case 1. Also, in Case 2, the derived equation 5x² + 3y² – 48 = 0 is not among the options, while 5x² + 3y² – 32 = 0 is different. Therefore, the correct equation is from Case 1.

Final Answer

The equation of the ellipse is $3x^2+5y^2-32=0$.

Related Topics and Formulae

Standard Equations of Ellipse

1. Major axis along x-axis: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, with $a>b$, eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$.

2. Major axis along y-axis: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, with $a>b$, eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$.

Key Points

• Eccentricity of an ellipse is between 0 and 1.
• The sum of distances from any point on the ellipse to the two foci is constant and equal to 2a.
• If the axes are along the coordinate axes, the center is at (0,0).