Find the equation to the ellipse with axes as the axes of coordinates.
distance between the foci is 10 and its latus rectum is 15,

Answer

To find the equation of the ellipse with the given conditions, we will follow these steps:

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

### Step 2: Set up the relationship for the foci
For an ellipse centered at the origin with the major axis along the x-axis, the foci are located at \((\pm ae, 0)\). The distance between the foci is given by \(2ae\). Therefore, we can set up the equation:
\[ 2ae = 10 \]
From this, we can solve for \(ae\):
\[ ae = 5 \quad \text{(Equation 1)} \]

### Step 3: Set up the relationship for the latus rectum
The length of the latus rectum \(L\) for an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
We know from the problem that the latus rectum is 15, so we can set up the equation:
\[ \frac{2b^2}{a} = 15 \]
From this, we can express \(b^2\) in terms of \(a\):
\[ 2b^2 = 15a \implies b^2 = \frac{15a}{2} \quad \text{(Equation 2)} \]

### Step 4: Use the relationship between \(a\), \(b\), and \(e\)
For an ellipse, the relationship between \(a\), \(b\), and \(e\) is given by:
\[ b^2 = a^2(1 - e^2) \]
Substituting \(b^2\) from Equation 2 into this equation gives:
\[ \frac{15a}{2} = a^2(1 - e^2) \]

### Step 5: Substitute \(e\) from Equation 1
From Equation 1, we know \(e = \frac{5}{a}\). Therefore, \(e^2 = \left(\frac{5}{a}\right)^2 = \frac{25}{a^2}\). Substituting this into the equation gives:
\[ \frac{15a}{2} = a^2\left(1 - \frac{25}{a^2}\right) \]
This simplifies to:
\[ \frac{15a}{2} = a^2 - 25 \]

### Step 6: Rearranging the equation
Rearranging this equation gives:
\[ a^2 - \frac{15a}{2} - 25 = 0 \]
To eliminate the fraction, multiply the entire equation by 2:
\[ 2a^2 - 15a - 50 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -15\), and \(c = -50\):
\[ a = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot (-50)}}{2 \cdot 2} \]
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 possible values for \(a\):
\[ a = \frac{40}{4} = 10 \quad \text{or} \quad a = \frac{-10}{4} = -2.5 \]
Since \(a\) must be positive, we take \(a = 10\).

### Step 8: Find \(b^2\)
Now substituting \(a = 10\) back into Equation 2:
\[ b^2 = \frac{15 \cdot 10}{2} = 75 \]

### Step 9: Write the equation of the ellipse
The standard form of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \(a^2 = 100\) and \(b^2 = 75\):
\[ \frac{x^2}{100} + \frac{y^2}{75} = 1 \]

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