Answer:
[tex]\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1[/tex]
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:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Let the point on the locus be P(x, y).
So, the distance from P to (0, 2) will be:
[tex]\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}[/tex]
And, the distance from P to (0, -2) will be:
[tex]\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}[/tex]
So sum of the two distances must be 6. Therefore:
[tex]d_1+d_2=6[/tex]
Now, by substitution:
[tex](\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6[/tex]
Simplify. We can subtract the second term from the left:
[tex]\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}[/tex]
Square both sides:
[tex](x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)[/tex]
We can cancel the x² terms and continue squaring:
[tex]y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4[/tex]
We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:
[tex]-8y=36-12\sqrt{x^2+(y+2)^2}[/tex]
We can divide both sides by -4:
[tex]2y=-9+3\sqrt{x^2+(y+2)^2}[/tex]
Adding 9 to both sides yields:
[tex]2y+9=3\sqrt{x^2+(y+2)^2}[/tex]
And, we will square both sides one final time.
[tex]4y^2+36y+81=9(x^2+(y^2+4y+4))[/tex]
Distribute:
[tex]4y^2+36y+81=9x^2+9y^2+36y+36[/tex]
The 36y will cancel. So:
[tex]4y^2+81=9x^2+9y^2+36[/tex]
Subtracting 4y² and 36 from both sides yields:
[tex]9x^2+5y^2=45[/tex]
And dividing both sides by 45 produces:
[tex]\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1[/tex]
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.
