The distance between the foci of an ellipse is 10 and its latus rectum is 15, find its equation referred to its axes as axes of coordinates.

Answer

To find the equation of the ellipse given the distance between the foci and the length of the latus rectum, we can follow these steps:

### Step 1: Understand the given information
We are given:
- The distance between the foci of the ellipse is 10.
- The length of the latus rectum is 15.

### Step 2: Set up the ellipse equation
Since the major axis is parallel to the x-axis, the standard form of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a > b\).

### Step 3: Relate the distance between the foci to \(a\) and \(e\)
The distance between the foci is given by \(2c\), where \(c = ae\) and \(e\) is the eccentricity. Therefore, we have:
\[ 2c = 10 \implies c = 5 \]
Thus, we have:
\[ ae = 5 \]

### Step 4: Use the latus rectum formula
The length of the latus rectum \(L\) for an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
We know that \(L = 15\), so:
\[ \frac{2b^2}{a} = 15 \implies 2b^2 = 15a \implies b^2 = \frac{15a}{2} \]

### Step 5: Relate \(e\) to \(a\) and \(b\)
The eccentricity \(e\) is also given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \(e = \frac{5}{a}\) into the equation gives:
\[ \frac{5}{a} = \sqrt{1 - \frac{b^2}{a^2}} \]

### Step 6: Substitute \(b^2\) into the eccentricity equation
Substituting \(b^2 = \frac{15a}{2}\) into the equation:
\[ \frac{5}{a} = \sqrt{1 - \frac{\frac{15a}{2}}{a^2}} \implies \frac{5}{a} = \sqrt{1 - \frac{15}{2a}} \]

### Step 7: Square both sides
Squaring both sides:
\[ \left(\frac{5}{a}\right)^2 = 1 - \frac{15}{2a} \]
This simplifies to:
\[ \frac{25}{a^2} = 1 - \frac{15}{2a} \]

### Step 8: Clear the fractions
Multiply through by \(2a^2\) to eliminate the denominators:
\[ 50 = 2a^2 - 15a \]
Rearranging gives:
\[ 2a^2 - 15a - 50 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ a = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot (-50)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ = \frac{15 \pm \sqrt{225 + 400}}{4} = \frac{15 \pm \sqrt{625}}{4} = \frac{15 \pm 25}{4} \]
This gives us two potential solutions:
\[ a = \frac{40}{4} = 10 \quad \text{and} \quad a = \frac{-10}{4} \text{ (not valid)} \]
Thus, \(a = 10\).

### Step 10: Find \(b^2\)
Now substituting \(a = 10\) back into the equation for \(b^2\):
\[ b^2 = \frac{15 \cdot 10}{2} = 75 \]

### Step 11: Write the equation of the ellipse
Substituting \(a^2\) and \(b^2\) into the standard form:
\[ \frac{x^2}{10^2} + \frac{y^2}{75} = 1 \]
This simplifies to:
\[ \frac{x^2}{100} + \frac{y^2}{75} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{100} + \frac{y^2}{75} = 1 \]