Cube root -512 Simplify.

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

jok3333 [9.3K]

Answer:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

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:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

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

$\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}$

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

$\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}$

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

$d_1+d_2=6$

Now, by substitution:

$(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6$

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

$\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$

Square both sides:

$(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)$

We can cancel the x² terms and continue squaring:

$y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4$

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

$-8y=36-12\sqrt{x^2+(y+2)^2}$

We can divide both sides by -4:

$2y=-9+3\sqrt{x^2+(y+2)^2}$

Adding 9 to both sides yields:

$2y+9=3\sqrt{x^2+(y+2)^2}$

And, we will square both sides one final time.

$4y^2+36y+81=9(x^2+(y^2+4y+4))$

Distribute:

$4y^2+36y+81=9x^2+9y^2+36y+36$

The 36y will cancel. So:

$4y^2+81=9x^2+9y^2+36$

Subtracting 4y² and 36 from both sides yields:

$9x^2+5y^2=45$

And dividing both sides by 45 produces:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

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.

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