Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6. ~Thanks in advance !

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:

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

And, the distance from P to (0, -2) will be:

So sum of the two distances must be 6. Therefore:

Now, by substitution:

Simplify. We can subtract the second term from the left:

Square both sides:

We can cancel the x² terms and continue squaring:

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

We can divide both sides by -4:

Adding 9 to both sides yields:

And, we will square both sides one final time.

Distribute:

The 36y will cancel. So:

Subtracting 4y² and 36 from both sides yields:

And dividing both sides by 45 produces:

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.

care to share were the trapezoid shown is ?

angle efh= angle egh