How do you find center, vertex, and foci of an ellipse $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ ?

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Answer 1 · 2016-02-15T09:54:59

Center of ellipse is $(2,-1)$ , vertices are $(-3, -1)$ and $(7, -1)$ and foci are $(-2, -1)$ and $(6, -1)$ .

The equation $9x^2+25y^2-36x+50y-164=0$ can be written as

$9(x^2-4x)+25(y^2+2y)=164$ or

$9(x^2-4x+4)+25(y^2+2y+1)=164+36+25$

$9(x-2)^2+25(y+1)^2=225$

Dividing both sides 225, we get

$9/225(x-2)^2+25/225(y+1)^2=1$ or $(x-2)^2/5^2+(y+1)^2/3^2=1$

Hence, Center of ellipse is $(2,-1)$ and major axis is $10$ ( $2*5$ ) and minor axis is $6$ ( $2*3$ ) and vertices are $(-3, -1)$ and $(7, -1)$ .

To find foci, $c^2=5^2-3^2=16=4^2$ . Hence foci are $(-2, -1)$ and $(6, -1)$ (4 units on major axis on either side of center).

How do you find center, vertex, and foci of an ellipse 9x^2 + 25y^2 - 36x + 50y - 164 = 09x^2 + 25y^2 - 36x + 50y - 164 = 0?