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the elliptical equation, ^2/16 + ^/4 = 1, the focci are located at pointsa. (±2√2, 0)b. (±2√3, 0)c. (±3√2, 0) d. (0, ±3√2) e. (4,2)

Answer :

Solution

We have the following equation:

[tex]\frac{x^2}{16}+\frac{y^2}{4}=1[/tex]

For this case the center is: C(0,0)

[tex]a=\sqrt[]{16}=4,b=\sqrt[]{4}=2[/tex]

We can find the value of c like this:

[tex]c=\sqrt[]{16-4}=\sqrt[]{12}=2\sqrt[]{3}[/tex]

Since x is the bigger axis we have the following coordinates for the focci:

[tex](0+2\sqrt[]{3},0),(0-2\sqrt[]{3},0)=(2\sqrt[]{3},0),(-2\sqrt[]{3},0)[/tex]

Then the answer is:

b. (±2√3, 0)

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