# What are the four vertices of (x+2)^2/9 +(y-3)^2/16 =1?

I already found two ((-2,7), (-2,-1)) with the formula but how do I find the other two?

### 8 Answers

- PinkgreenLv 72 months ago
Let x+2=0, then [(y-3)^2]/16=1=>y-3=+/-4=>

y=3+4=7 or y=3-4=-1

=> (-2,7) & (-2,-1) are 2 vertices.

Let y-3=0, then [(x+2)^2]/9=1=>x+2=+/-3=>

x=-2+3=1 or x-2-3=-5

=> (1,3) & (-5,3) are 2 vertices.

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- Φ² = Φ+1Lv 72 months ago
You have already found two: (-2,7) and (-2,-1).

They are √16 = 4 units from their midpoint, which is (-2,3).

The other two vertices are √9 = 3 units away from this point and lie on the perpendicular bisector of the first two points, so (-1,3) and (5,3).

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- DWReadLv 72 months ago
(x+2)²/9 + (y-3)²/16 = 1 is the equation of a vertical ellipse.

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- Engr. RonaldLv 72 months ago
(x + 2)^2..... (y - 3)^2

------------ + -------------- = 1

... 9.................. 16

from this formula

(x - h)^2..(y - k)^2

--------- + ----------- = 1

...b^2......... a^2

finding other vertices of the ellipse

given our center of ellipse is (- 2,3) and a = 3 and b = 4

Now use the formula from the coordinates of the minor axis on the vertical ellipse o

on the left side.

(h - a,k) = (-2 -3, 3) = (-5, 3) answer//

(h + a, k) = (-2 + 3, 3) = (1, 3) answer//

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- sepiaLv 72 months ago
((x + 2)^2)/9 + ((y - 3)^2)/16 = 1

((-2,7), (-2,-1), (-5, 3), (1, 3))

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- davidLv 72 months ago
y = 3

... ( x+2)^2 = 9

x + 2 = +/- 3

x = -5 and x = 1

(-5, 3) and (1, 3)

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- 2 months ago
Centered at (-2 , 3)

Semi-minor axis is 3 units in length and is parallel to the x-axis

Semi-minor axis is 4 units in length and is parallel to the y-axis

(-2 - 3 , 3)

(-2 + 3 , 3)

(-2 , 3 - 4)

(-2 , 3 + 4)

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