To find the equation of the locus of a point such that the sum of its distances from the points \((0,2)\) and \((0,-2)\) is 6 units, we begin by noting that such a locus describes an ellipse. Here’s a structured step-by-step solution:
1. Identify the Foci and Sum of Distances:
- The points \((0, 2)\) and \((0, -2)\) are the foci of the ellipse.
- The sum of distances from any point on the ellipse to these two fixed points (foci) is given as 6 units.
2. Determine the Semi-Major Axis (a):
- In an ellipse, the sum of distances from any point on the ellipse to the foci is equal to the length of the major axis, which is \(2a\).
- Given sum of distances is \(6\) units, so \(2a = 6\).
- Thus, the semi-major axis, \(a\), is \(a = \frac{6}{2} = 3\).
3. Calculate the Distance Between the Foci (2c):
- The distance between the foci \((0, 2)\) and \((0, -2)\) is calculated as \(2c\).
- The distance between the points \((0, 2)\) and \((0, -2)\) is \(4\) units.
- Hence, \(2c = 4\) which gives \(c = 2\).
4. Determine the Semi-Minor Axis (b):
- Using the relationship for ellipses: \(c^2 = a^2 - b^2\), we can solve for \(b\).
- Here, we have \(a = 3\) and \(c = 2\).
- Substitute these values into the equation:
[tex]\[
c^2 = a^2 - b^2
\][/tex]
[tex]\[
2^2 = 3^2 - b^2
\][/tex]
[tex]\[
4 = 9 - b^2
\][/tex]
[tex]\[
b^2 = 9 - 4
\][/tex]
[tex]\[
b^2 = 5
\][/tex]
- Therefore, \(b = \sqrt{5} \approx 2.236\).
5. Formulate the Equation of the Ellipse:
- The standard form of the equation of an ellipse centered at the origin \((0,0)\) with semi-major axis \(a\) and semi-minor axis \(b\) is:
[tex]\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\][/tex]
- Plugging in the values of \(a\) and \(b\):
[tex]\[
\frac{x^2}{3^2} + \frac{y^2}{(\sqrt{5})^2} = 1
\][/tex]
[tex]\[
\frac{x^2}{9} + \frac{y^2}{5} = 1
\][/tex]
Therefore, the equation of the locus of a point such that the sum of its distances from \((0,2)\) and \((0,-2)\) is 6 units is:
[tex]\[
\boxed{\frac{x^2}{9} + \frac{y^2}{5} = 1}
\][/tex]
