Answer:
![\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bx%5E2%7D%7B5%7D%2B%5Cfrac%7By%5E2%7D%7B9%7D%3D1)
Step-by-step explanation:
We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.
First, we will need the distance formula, given by:
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Let the point on the locus be P(x, y).
So, the distance from P to (0, 2) will be:
![\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20d_1%26%3D%5Csqrt%7B%28x-0%29%5E2%2B%28y-2%29%5E2%7D%5C%5C%5C%5C%20%26%3D%5Csqrt%7Bx%5E2%2B%28y-2%29%5E2%7D%5Cend%7Baligned%7D)
And, the distance from P to (0, -2) will be:
![\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20d_2%26%3D%5Csqrt%7B%28x-0%29%5E2%2B%28y-%28-2%29%29%5E2%7D%5C%5C%5C%5C%20%26%3D%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D%5Cend%7Baligned%7D)
So sum of the two distances must be 6. Therefore:
![d_1+d_2=6](https://tex.z-dn.net/?f=d_1%2Bd_2%3D6)
Now, by substitution:
![(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6](https://tex.z-dn.net/?f=%28%5Csqrt%7Bx%5E2%2B%28y-2%29%5E2%7D%29%2B%28%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D%29%3D6)
Simplify. We can subtract the second term from the left:
![\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%2B%28y-2%29%5E2%7D%3D6-%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D)
Square both sides:
![(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)](https://tex.z-dn.net/?f=%28x%5E2%2B%28y-2%29%5E2%29%3D36-12%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D%2B%28x%5E2%2B%28y%2B2%29%5E2%29)
We can cancel the x² terms and continue squaring:
![y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4](https://tex.z-dn.net/?f=y%5E2-4y%2B4%3D36-12%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D%2By%5E2%2B4y%2B4)
We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:
![-8y=36-12\sqrt{x^2+(y+2)^2}](https://tex.z-dn.net/?f=-8y%3D36-12%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D)
We can divide both sides by -4:
![2y=-9+3\sqrt{x^2+(y+2)^2}](https://tex.z-dn.net/?f=2y%3D-9%2B3%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D)
Adding 9 to both sides yields:
![2y+9=3\sqrt{x^2+(y+2)^2}](https://tex.z-dn.net/?f=2y%2B9%3D3%5Csqrt%7Bx%5E2%2B%28y%2B2%29%5E2%7D)
And, we will square both sides one final time.
![4y^2+36y+81=9(x^2+(y^2+4y+4))](https://tex.z-dn.net/?f=4y%5E2%2B36y%2B81%3D9%28x%5E2%2B%28y%5E2%2B4y%2B4%29%29)
Distribute:
![4y^2+36y+81=9x^2+9y^2+36y+36](https://tex.z-dn.net/?f=4y%5E2%2B36y%2B81%3D9x%5E2%2B9y%5E2%2B36y%2B36)
The 36y will cancel. So:
![4y^2+81=9x^2+9y^2+36](https://tex.z-dn.net/?f=4y%5E2%2B81%3D9x%5E2%2B9y%5E2%2B36)
Subtracting 4y² and 36 from both sides yields:
![9x^2+5y^2=45](https://tex.z-dn.net/?f=9x%5E2%2B5y%5E2%3D45)
And dividing both sides by 45 produces:
![\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bx%5E2%7D%7B5%7D%2B%5Cfrac%7By%5E2%7D%7B9%7D%3D1)
Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.