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9x2+4y2 = 36 The foci are located at:

User Allen Chan
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1 Answer

2 votes

Answer:

The foci are located at
(0,\pm √(5))

Explanation:

The standard equation of an ellipse with a vertical major axis is
(x^2)/(b^2)+(y^2)/(a^2)=1 where
a^2\:>\:b^2.

The given equation is
9x^2+4y^2=36.

To obtain the standard form, we must divide through by 36.


(9x^2)/(36)+(4y^2)/(36)=(36)/(36)

We simplify by canceling out the common factors to obtain;


(x^2)/(4) +(y^2)/(9)=1

By comparing this equation to


(x^2)/(b^2)+(y^2)/(a^2)=1

We have
a^2=9,b^2=4.

To find the foci, we use the relation:
a^2-b^2=c^2

This implies that:


9-4=c^2


c^2=5


c=\pm√(5)

The foci are located at
(0,\pm c)

Therefore the foci are
(0,\pm √(5))

Or


(0,-√(5)) and
(0,√(5))

User ZiiMakc
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