A point P lies on the ellipe `((y-1)^(2))/(64)+((x+2)^(2))/(49)=1` . If the distance of P from one focus is 10 units, then find its distance from other focus.

Explanation

To solve the problem step by step, we will analyze the given ellipse and use the properties of ellipses to find the required distance.

### Step 1: Identify the equation of the ellipse
The given equation of the ellipse is:
\[ \frac{(y-1)^2}{64} + \frac{(x+2)^2}{49} = 1 \]
This can be rewritten as:
\[ \frac{(x+2)^2}{7^2} + \frac{(y-1)^2}{8^2} = 1 \]
Here, we can identify \(a = 7\) and \(b = 8\).

### Step 2: Determine the foci of the ellipse
For an ellipse, the distance of the foci from the center is given by:
\[ c = \sqrt{b^2 - a^2} \]
Substituting the values of \(a\) and \(b\):
\[ c = \sqrt{8^2 - 7^2} = \sqrt{64 - 49} = \sqrt{15} \]
Thus, the foci are located at:
\[ (-2, 1 + \sqrt{15}) \quad \text{and} \quad (-2, 1 - \sqrt{15}) \]

### Step 3: Use the property of the ellipse
One of the properties of an ellipse states that for any point \(P\) on the ellipse, the sum of the distances from \(P\) to the two foci is constant and equal to \(2b\):
\[ Pf + Pf' = 2b \]
Here, \(b = 8\), so:
\[ Pf + Pf' = 2 \times 8 = 16 \]

### Step 4: Given distance from one focus
According to the problem, the distance from point \(P\) to one of the foci (let's say \(Pf\)) is given as 10 units:
\[ Pf = 10 \]

### Step 5: Find the distance from the other focus
Using the property of the ellipse:
\[ Pf' = 16 - Pf \]
Substituting \(Pf = 10\):
\[ Pf' = 16 - 10 = 6 \]

### Conclusion
The distance from point \(P\) to the other focus \(Pf'\) is:
\[ \boxed{6} \]

---