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 :

Options

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

Answer

To find the equation of the ellipse whose axes are the coordinate axes, passes through the point (-3, 1), and has an eccentricity of \(\sqrt{\frac{2}{5}}\), we will follow these steps:

### Step 1: Write the standard form of the ellipse
The standard form of the ellipse centered at the origin with axes along the coordinate axes is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

### Step 2: Use the eccentricity formula
The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Given that \(e = \sqrt{\frac{2}{5}}\), we can square both sides:
\[ e^2 = \frac{2}{5} = 1 - \frac{b^2}{a^2} \]
Rearranging gives:
\[ \frac{b^2}{a^2} = 1 - \frac{2}{5} = \frac{3}{5} \]

### Step 3: Express \(b^2\) in terms of \(a^2\)
From the equation \(\frac{b^2}{a^2} = \frac{3}{5}\), we can express \(b^2\) as:
\[ b^2 = \frac{3}{5} a^2 \]

### Step 4: Substitute \(b^2\) into the ellipse equation
Substituting \(b^2\) into the standard form of the ellipse:
\[ \frac{x^2}{a^2} + \frac{y^2}{\frac{3}{5}a^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{a^2} + \frac{5y^2}{3a^2} = 1 \]
Multiplying through by \(3a^2\) gives:
\[ 3x^2 + 5y^2 = 3a^2 \]

### Step 5: Use the point (-3, 1) to find \(a^2\)
Now we substitute the point (-3, 1) into the equation:
\[ 3(-3)^2 + 5(1)^2 = 3a^2 \]
Calculating the left side:
\[ 3(9) + 5(1) = 27 + 5 = 32 \]
Thus, we have:
\[ 32 = 3a^2 \implies a^2 = \frac{32}{3} \]

### Step 6: Find \(b^2\)
Now substituting \(a^2\) back to find \(b^2\):
\[ b^2 = \frac{3}{5} a^2 = \frac{3}{5} \cdot \frac{32}{3} = \frac{32}{5} \]

### Step 7: Write the final equation of the ellipse
Substituting \(a^2\) and \(b^2\) back into the ellipse equation:
\[ 3x^2 + 5y^2 = 32 \]
This can be rearranged to:
\[ 3x^2 + 5y^2 - 32 = 0 \]

### Final Answer
The equation of the ellipse is:
\[ 3x^2 + 5y^2 - 32 = 0 \]