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 ! ​

Answers

Answer 1

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.


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Answer:

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Step-by-step explanation:

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10
9
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5
4
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Answers

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Answers

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Answers

Answer:

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Answers

Answer:

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Step-by-step explanation:

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2

Step-by-step explanation:

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Answers

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Answer:

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Is this what you're asking?

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