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 centered at the origin and passing through the points (-3, 1) and (2, -2), we can start with the standard form of the equation of an ellipse:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

### Step 1: Substitute the first point (-3, 1) into the ellipse equation.

Substituting \(x = -3\) and \(y = 1\):

\[ \frac{(-3)^2}{a^2} + \frac{(1)^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{9}{a^2} + \frac{1}{b^2} = 1 \quad \text{(Equation 1)} \]

### Step 2: Substitute the second point (2, -2) into the ellipse equation.

Now, substituting \(x = 2\) and \(y = -2\):

\[ \frac{(2)^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{4}{a^2} + \frac{4}{b^2} = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations.

Now we have two equations:

1. \(\frac{9}{a^2} + \frac{1}{b^2} = 1\) (Equation 1)
2. \(\frac{4}{a^2} + \frac{4}{b^2} = 1\) (Equation 2)

Let's express \(\frac{1}{b^2}\) from Equation 1:

\[ \frac{1}{b^2} = 1 - \frac{9}{a^2} \]

Substituting this into Equation 2:

\[ \frac{4}{a^2} + 4\left(1 - \frac{9}{a^2}\right) = 1 \]

This simplifies to:

\[ \frac{4}{a^2} + 4 - \frac{36}{a^2} = 1 \]

Combining like terms gives:

\[ -\frac{32}{a^2} + 4 = 1 \]

Subtracting 4 from both sides:

\[ -\frac{32}{a^2} = -3 \]

Multiplying both sides by -1:

\[ \frac{32}{a^2} = 3 \]

Cross-multiplying gives:

\[ 32 = 3a^2 \implies a^2 = \frac{32}{3} \]

### Step 4: Find \(b^2\).

Now substitute \(a^2\) back into Equation 1 to find \(b^2\):

\[ \frac{9}{\frac{32}{3}} + \frac{1}{b^2} = 1 \]

This simplifies to:

\[ \frac{27}{32} + \frac{1}{b^2} = 1 \]

Subtracting \(\frac{27}{32}\) from both sides:

\[ \frac{1}{b^2} = 1 - \frac{27}{32} = \frac{32 - 27}{32} = \frac{5}{32} \]

Taking the reciprocal gives:

\[ b^2 = \frac{32}{5} \]

### Step 5: Write the final equation of the ellipse.

Now we have \(a^2 = \frac{32}{3}\) and \(b^2 = \frac{32}{5}\). Therefore, the equation of the ellipse is:

\[ \frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}} = 1 \]

Multiplying through by \( \frac{32}{15} \) to eliminate the denominators gives:

\[ \frac{15x^2}{32} + \frac{15y^2}{32} = 1 \]

Thus, the final equation of the ellipse is:

\[ \frac{15x^2}{32} + \frac{15y^2}{32} = 1 \]