Find the vertices of the ellipse defined by the equation shown below. If necessary, round to the nearest tenth. 25x^(2)+4y^(2)-200x+8y+340=0
To find the vertices of the ellipse, we need to rewrite the equation in standard form:
25x^(2) + 4y^(2) - 200x + 8y + 340 = 0
Divide both sides by 25:
x^(2) - 8x + (4/25)y^(2) + 2y + 34/5 = 0
Complete the square for x:
x^(2) - 8x + 16 + (4/25)y^(2) + 2y + 34/5 = 16
Simplify:
(1/25)(4y^(2) + 100y + 666) + (1/25)(x - 4)^(2) = 1
Now we can see that this is an ellipse with center (4, -25/2), semi-major axis length of 5, and semi-minor axis length of sqrt(25/4) = 5/2.
The vertices are located at (4, -25/2 ± 5).
Therefore, the vertices are:
(4, -22.5) and (4, -27.5)
Find the vertices of the ellipse defined by the equation shown below. If necessary, round to the nearest tenth.
16x^2 + 9y^2 +96x-36y+36= 0
FInd what the Foci is
To find the vertices and foci of the ellipse, we need to rewrite the equation in standard form:
16x^2 + 9y^2 +96x-36y+36= 0
Divide both sides by 144:
x^2 + 6x/9 + y^2/16 - 1/4 = 0
Complete the square for x:
x^2 + 6x/9 + 1/9 + y^2/16 - 1/4 = 1/9
Simplify:
(x + 1/3)^2/1/9 + y^2/16 = 1
Now we can see that this is an ellipse with center (-1/3,0), semi-major axis length of 3, and semi-minor axis length of 4.
The vertices are located at (-1/3 ± 3, 0).
Therefore, the vertices are:
(-4.333, 0) and (4.333, 0)
To find the foci, we need to use the formula:
c^2 = a^2 - b^2
where a = 3 and b = 4.
c^2 = 9 - 16
c^2 = -7 (this means that the ellipse is imaginary and has no real foci)
there is a foci
If there is a foci, then the ellipse is actually a hyperbola with a center (h,k). We can use the standard form of a hyperbola to find the foci:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
Comparing this to the equation we have, we can see that the x term has a coefficient of 16 and the y term has a coefficient of 9. Since the x term has the larger coefficient, we know that the hyperbola will have a horizontal transverse axis.
To find the center, we need to complete the square:
16x^2 + 9y^2 + 96x - 36y + 36 = 0
16(x^2 + 6x) + 9(y^2 - 4y) = -36
16(x^2 + 6x + 9) + 9(y^2 - 4y + 4) = -36 + 144 + 36
16(x + 3)^2 + 9(y - 2)^2 = 144
Dividing by 144, we get:
(x + 3)^2/9 + (y - 2)^2/16 = 1
So the center is (-3, 2).
The equation is of the form
(x + 3)^2/9 - (y - 2)^2/16 = 1
so we can identify a^2 = 9 and b^2 = 16, and therefore c^2 = a^2 + b^2 = 25.
The distance from the center to each focus is c, or 5.
Therefore, the foci are located at:
(-3 - 5, 2) and (-3 + 5, 2)
which simplify to
(-8, 2) and (2, 2)
your answer was wrong the answer is Foci: (−3,4.6) and (−3,−0.6)
I apologize for the mistake in my previous answer. Thank you for letting me know.
To actually find the foci, we can use the formula:
c^2 = a^2 + b^2
where a is the semi-major axis and b is the semi-minor axis. From our standard form, we can see that a^2 = 9 and b^2 = 16, so:
c^2 = 9 + 16
c^2 = 25
c = 5
So the distance from the center to each foci is 5. Since the transverse axis of the hyperbola is horizontal, the foci will be located on the line y = k. The center of the hyperbola is (-3,0), so the foci will be at (-3, ±5).
However, the original equation is in vertex form, not standard form. So we need to first rewrite the equation in the standard form, before we can find the vertices and foci. To do this, we first need to factor out the coefficient of x^2 and y^2:
16x^2 + 9y^2 + 96x - 36y + 36 = 0
16(x^2 + 6x) + 9(y^2 - 4y) = -36
16(x^2 + 6x + 9) + 9(y^2 - 4y + 4) = -36 + 144 + 36
16(x + 3)^2 + 9(y - 2)^2 = 144
(x + 3)^2/9 + (y - 2)^2/16 = 1
Now we can see that the standard form of the hyperbola is:
(x + 3)^2/9 - (y - 2)^2/16 = 1
The center of the hyperbola is (-3, 2), so the foci will be at (-3, 2 ± 5):
(-3, 2 + 5) = (-3, 7)
(-3, 2 - 5) = (-3, -3)
So the foci are located at:
(-3, 7) and (-3, -3)
I hope this answer is correct. Please let me know if you have any questions or concerns.