Find the equation of the ellipse whose foci are `(pm3,0)` and it passes through the point `(2,sqrt(7))`.

Answer

To find the equation of the ellipse whose foci are at (±3, 0) and which passes through the point (2, √7), we will follow these steps:

### Step 1: Identify the values of a and c
The foci of the ellipse are given as (±3, 0). In an ellipse, the distance from the center to each focus is denoted as c. Therefore, we have:
\[ c = 3 \]

### Step 2: Relate a, b, and c
For an ellipse, the relationship between a (semi-major axis), b (semi-minor axis), and c is given by:
\[ c^2 = a^2 - b^2 \]

### Step 3: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 4: Use the point (2, √7) to find a and b
Since the ellipse passes through the point (2, √7), we can substitute these values into the ellipse equation:
\[ \frac{2^2}{a^2} + \frac{(\sqrt{7})^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} + \frac{7}{b^2} = 1 \]

### Step 5: Express b^2 in terms of a^2
From the relationship \( c^2 = a^2 - b^2 \), we can express b^2 as:
\[ b^2 = a^2 - c^2 = a^2 - 9 \]

### Step 6: Substitute b^2 into the equation
Now substitute \( b^2 = a^2 - 9 \) into the equation we derived from the point:
\[ \frac{4}{a^2} + \frac{7}{a^2 - 9} = 1 \]

### Step 7: Solve for a^2
To solve for a^2, we can find a common denominator and rearrange the equation:
\[ \frac{4(a^2 - 9) + 7a^2}{a^2(a^2 - 9)} = 1 \]
This leads to:
\[ 4a^2 - 36 + 7a^2 = a^2(a^2 - 9) \]
Combining terms gives:
\[ 11a^2 - 36 = a^4 - 9a^2 \]
Rearranging this gives:
\[ a^4 - 20a^2 + 36 = 0 \]

### Step 8: Let \( t = a^2 \)
Letting \( t = a^2 \), we can rewrite the equation as:
\[ t^2 - 20t + 36 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1} \]
\[ t = \frac{20 \pm \sqrt{400 - 144}}{2} \]
\[ t = \frac{20 \pm \sqrt{256}}{2} \]
\[ t = \frac{20 \pm 16}{2} \]
This gives us:
\[ t = 18 \quad \text{or} \quad t = 2 \]

### Step 10: Determine valid a^2 and b^2
Since \( a^2 = 18 \) and \( a^2 = 2 \) are possible, we need to check which one is valid:
1. If \( a^2 = 18 \), then \( b^2 = 18 - 9 = 9 \).
2. If \( a^2 = 2 \), then \( b^2 = 2 - 9 = -7 \) (not valid).

Thus, we have:
\[ a^2 = 18 \quad \text{and} \quad b^2 = 9 \]

### Step 11: Write the final equation of the ellipse
Substituting these values back into the standard form gives:
\[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \]

### Final Answer:
The equation of the ellipse is:
\[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \]

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