I am supposed to find the 4 vertices of the ellipse
1=(((x+1)^2)/4)+(((y-4)^20/9)
How would I go about doing that?
in
1=(((x+1)^2)/4)+(((y-4)^2)/9) (I fixed your typo) or
9(x+1)^2 + 4(y-4)^2 = 36
let x=0
then 9 + 4(y-4)^2 = 36
(y-4)^2 = 27/4
y-4 = ± (3√3)/4
so the y-intercepts are
4 + (3√3)/4 and 4 - (3√3)/4
let y = 0 to find the x-intercepts in the same way
My teacher told me the answers were (-1,7) (-3,4) (1,4) and (-1,1). But i just didn't understand how to do it, the bell rang.
I misread your question.
I found the intercept, but you did not ask for these.
ok
had it been (x^2)/4 + (y^2)/9 = 1
the centre would have been (0,0)
the vertices would have been (2,0), (-2,0), (0,3) and (0,-3)
Your equation has the same shape , but its centre is at (-1,4)
this ellipse has been translated 1 unit to the left, and 4 units up.
so the vertices are
(2,0) ---> (1,4)
(-2,0( ---> (-3,4)
(0,3) ---> (-1,7)
(0,-3) --> (-1,1)
those match your teacher's answers
To find the vertices of an ellipse, you need to know the center, semi-major axis, and semi-minor axis of the ellipse. The equation provided is in a standard form for an ellipse:
1 = ((x + h)²) / a² + ((y - k)²) / b²
Where (h, k) is the center of the ellipse, a is the semi-major axis, and b is the semi-minor axis.
Comparing this equation with the given equation:
1 = ((x + 1)²) / 4 + ((y - 4)²) / 9
we can see that:
Center: (-1, 4)
Semi-major axis: a = 2 (square root of 4)
Semi-minor axis: b = 3 (square root of 9)
To find the vertices, you need to consider the distance of a (the semi-major axis) from the center along the x-axis. Since the center is at (-1, 4), the vertices will be at (-1 - a, 4) and (-1 + a, 4).
Using these values, the vertices of the ellipse are:
Vertex 1: (-1 - 2, 4) = (-3, 4)
Vertex 2: (-1 + 2, 4) = (1, 4)
So, the four vertices of the ellipse are (-3, 4), (1, 4), (-1, 1), and (-1, 7).