The equation of the ellipse whose centre is at origin and which passes through the points (-3,1) and (2,-2) is

Options

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

Answer

To find the equation of the ellipse whose center is at the origin and which passes through the points (-3, 1) and (2, -2), we can follow these steps:

### Step 1: Write the standard equation of the ellipse
The standard equation of an ellipse centered at the origin is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 2: Substitute the first point (-3, 1)
Substituting the point (-3, 1) into the equation:
\[ \frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} + \frac{1}{b^2} = 1 \]
Multiplying through by \(a^2b^2\) to eliminate the denominators gives:
\[ 9b^2 + a^2 = a^2b^2 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (2, -2)
Now, substitute the point (2, -2) into the equation:
\[ \frac{(2)^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} + \frac{4}{b^2} = 1 \]
Multiplying through by \(a^2b^2\) gives:
\[ 4b^2 + 4a^2 = a^2b^2 \quad \text{(Equation 2)} \]

### Step 4: Set up the equations
From Equation 1:
\[ 9b^2 + a^2 = a^2b^2 \]
From Equation 2:
\[ 4b^2 + 4a^2 = a^2b^2 \]

### Step 5: Equate the two equations
From Equation 1, we can rearrange it to:
\[ a^2b^2 - 9b^2 - a^2 = 0 \]
From Equation 2, we can rearrange it to:
\[ a^2b^2 - 4b^2 - 4a^2 = 0 \]

### Step 6: Solve for \(a^2\) and \(b^2\)
Setting the two expressions for \(a^2b^2\) equal to each other:
\[ 9b^2 + a^2 = 4b^2 + 4a^2 \]
Rearranging gives:
\[ 5b^2 = 3a^2 \]
Thus,
\[ a^2 = \frac{5}{3}b^2 \]

### Step 7: Substitute back into one of the equations
Substituting \(a^2\) back into Equation 1:
\[ 9b^2 + \frac{5}{3}b^2 = \frac{5}{3}b^2b^2 \]
Multiplying through by 3 to eliminate the fraction:
\[ 27b^2 + 5b^2 = 5b^4 \]
This simplifies to:
\[ 32b^2 = 5b^4 \]
Dividing both sides by \(b^2\) (assuming \(b \neq 0\)):
\[ 32 = 5b^2 \]
Thus,
\[ b^2 = \frac{32}{5} \]

### Step 8: Find \(a^2\)
Now substituting \(b^2\) back to find \(a^2\):
\[ a^2 = \frac{5}{3} \cdot \frac{32}{5} = \frac{32}{3} \]

### Step 9: Write the final equation of the ellipse
Substituting \(a^2\) and \(b^2\) back into the standard form:
\[ \frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}} = 1 \]
Multiplying through by 15 (the least common multiple of the denominators 3 and 5):
\[ 15 \left(\frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}}\right) = 15 \]
This simplifies to:
\[ \frac{45x^2}{32} + \frac{75y^2}{32} = 15 \]
Multiplying through by 32 gives:
\[ 45x^2 + 75y^2 = 480 \]
Dividing through by 15 gives:
\[ 3x^2 + 5y^2 = 32 \]

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