Writing Ellipses Equations worksheet with ten problems involving centers, vertices, foci, and widths to determine standard form equations.

Final Answer:
1) $\frac{(x - 6)^2}{36} + \frac{(y + 8)^2}{20} = 1$
2) $\frac{(x + 1)^2}{4} + \frac{(y - 6)^2}{56} = 1$
3) $\frac{x^2}{36} + \frac{(y + 5)^2}{25} = 1$
4) $\frac{(x + 4)^2}{16} + \frac{(y - 5)^2}{48} = 1$
5) $\frac{(x - 5)^2}{9} + \frac{(y - 6)^2}{49} = 1$
6) $\frac{(x - 2)^2}{144} + \frac{(y + 7)^2}{128} = 1$
7) $\frac{(x - 6)^2}{16} + \frac{(y + 7)^2}{16} = 1$ → Wait, this is a circle? No — vertices at (9,−7) and (−3,−7) ⇒ horizontal major axis, center at ((9−3)/2, −7) = (3, −7), so *a* = distance from center to vertex = 6 ⇒ $a^2 = 36$. Co-vertices at (3,−4) and (3,−10) ⇒ vertical minor axis, *b* = 3 ⇒ $b^2 = 9$. So correct: $\frac{(x - 3)^2}{36} + \frac{(y + 7)^2}{9} = 1$
8) Foci at $(2 \pm \sqrt{95}, 6)$, minor axis endpoints at $(2,13)$ and $(2,-1)$ ⇒ center at (2,6), *b* = distance from center to minor endpoint = 7 ⇒ $b^2 = 49$. Distance from center to focus = $c = \sqrt{95}$, and $c^2 = a^2 - b^2$ ⇒ $a^2 = c^2 + b^2 = 95 + 49 = 144$. Major axis is horizontal (foci left/right), so: $\frac{(x - 2)^2}{144} + \frac{(y - 6)^2}{49} = 1$
9) Foci at $(6 \pm \sqrt{17}, -8)$, co-vertices at $(6,0)$ and $(6,-16)$ ⇒ center at (6,−8), *b* = 8 ⇒ $b^2 = 64$. $c = \sqrt{17}$, so $a^2 = c^2 + b^2 = 17 + 64 = 81$. Horizontal major axis: $\frac{(x - 6)^2}{81} + \frac{(y + 8)^2}{64} = 1$
10) Center (−2,−6), vertex (7,−6) ⇒ horizontal major axis, *a* = distance = 9 ⇒ $a^2 = 81$. Focus at $(−2 + 2\sqrt{14}, −6)$ ⇒ $c = 2\sqrt{14}$, so $c^2 = 56$. Then $b^2 = a^2 - c^2 = 81 - 56 = 25$. Equation: $\frac{(x + 2)^2}{81} + \frac{(y + 6)^2}{25} = 1$

Let me double-check each carefully:

1) Center (6,−8), vertex (12,−8) ⇒ horizontal, *a* = 6 ⇒ $a^2=36$. Co-vertex (6,−4) ⇒ vertical, *b* = 4 ⇒ $b^2=16$? Wait! Mistake: center (6,−8), co-vertex (6,−4) ⇒ distance = |−4 − (−8)| = 4 ⇒ $b = 4$, so $b^2 = 16$, not 20. But check: for ellipse, $c^2 = a^2 - b^2$. Focus given? Not directly, but we can verify: if a=6, b=4, then c = √(36−16)=√20=2√5. But no focus given in #1, so use only given points. Vertex and co-vertex define a and b directly. So yes: $a=6$, $b=4$ ⇒ equation: $\frac{(x-6)^2}{36} + \frac{(y+8)^2}{16} = 1$

I made an error earlier. Let's redo all with care:

General standard form:
- Horizontal major axis: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $a > b$, center $(h,k)$, vertices at $(h\pm a, k)$, co-vertices at $(h, k\pm b)$, foci at $(h\pm c, k)$, $c^2 = a^2 - b^2$
- Vertical major axis: $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$, $a > b$, vertices at $(h, k\pm a)$, co-vertices at $(h\pm b, k)$

Now solve each:

1) Center (6,−8), Vertex (12,−8) → same y, so horizontal major axis. Distance = 12−6 = 6 ⇒ $a = 6$. Co-vertex (6,−4) → same x, distance = |−4 − (−8)| = 4 ⇒ $b = 4$. So:
$\frac{(x-6)^2}{36} + \frac{(y+8)^2}{16} = 1$

2) Vertices (−1,15), (−1,−3) → same x, vertical major axis. Center = midpoint = (−1, (15 + (−3))/2) = (−1, 6). $a =$ distance from center to vertex = |15−6| = 9 ⇒ $a^2 = 81$. Foci given: (−1, 6 ± 2√14) ⇒ $c = 2\sqrt{14}$, $c^2 = 56$. Then $b^2 = a^2 - c^2 = 81 − 56 = 25$. So vertical:
$\frac{(x+1)^2}{25} + \frac{(y-6)^2}{81} = 1$

3) Center (0,−5), Width = 12 ⇒ width = 2b if vertical? Wait: “Width” usually means horizontal span → total length of minor axis if major is vertical, or major axis if horizontal. But since center given and width = 12, and focus has same y as center (−5), focus at (√35, −5) ⇒ horizontal displacement ⇒ major axis is horizontal. So width = 2a = 12 ⇒ $a = 6$, $a^2 = 36$. Focus at (√35, −5) ⇒ $c = \sqrt{35}$, so $c^2 = 35$. Then $b^2 = a^2 - c^2 = 36 − 35 = 1$. So:
$\frac{x^2}{36} + \frac{(y+5)^2}{1} = 1$

4) Foci (−4, 5 ± 4√7) ⇒ vertical line x = −4, so major axis vertical. Center = (−4, 5). $c = 4\sqrt{7}$, $c^2 = 112$. Major axis endpoints: (−4,16) and (−4,−6) ⇒ distance from center to endpoint = |16−5| = 11 ⇒ $a = 11$, $a^2 = 121$. Then $b^2 = a^2 - c^2 = 121 − 112 = 9$. So vertical ellipse:
$\frac{(x+4)^2}{9} + \frac{(y-5)^2}{121} = 1$

5) Center (5,6), Vertex (5,13) ⇒ same x, vertical major axis. $a = |13−6| = 7$, $a^2 = 49$. Co-vertex (8,6) ⇒ same y, $b = |8−5| = 3$, $b^2 = 9$. So:
$\frac{(x-5)^2}{9} + \frac{(y-6)^2}{49} = 1$

6) Center (2,−7), Width = 24 ⇒ if width is horizontal span, and focus has same y, likely horizontal major axis. So 2a = 24 ⇒ $a = 12$, $a^2 = 144$. Focus at $(2 + 8\sqrt{2}, -7)$ ⇒ $c = 8\sqrt{2}$, $c^2 = 128$. Then $b^2 = a^2 - c^2 = 144 − 128 = 16$. So:
$\frac{(x-2)^2}{144} + \frac{(y+7)^2}{16} = 1$

7) Vertices (9,−7), (−3,−7) → horizontal, center = ((9−3)/2, −7) = (3, −7), $a = 6$, $a^2 = 36$. Co-vertices (3,−4), (3,−10) → vertical, $b = 3$, $b^2 = 9$. So:
$\frac{(x-3)^2}{36} + \frac{(y+7)^2}{9} = 1$

8) Foci $(2 \pm \sqrt{95}, 6)$ ⇒ horizontal, center (2,6), $c^2 = 95$. Minor axis endpoints (2,13), (2,−1) ⇒ vertical, so $b = |13−6| = 7$, $b^2 = 49$. Then $a^2 = b^2 + c^2 = 49 + 95 = 144$. So:
$\frac{(x-2)^2}{144} + \frac{(y-6)^2}{49} = 1$

9) Foci $(6 \pm \sqrt{17}, -8)$ ⇒ horizontal, center (6,−8), $c^2 = 17$. Co-vertices (6,0), (6,−16) ⇒ vertical, $b = |0 − (−8)| = 8$, $b^2 = 64$. So $a^2 = b^2 + c^2 = 64 + 17 = 81$. Equation:
$\frac{(x-6)^2}{81} + \frac{(y+8)^2}{64} = 1$

10) Center (−2,−6), Vertex (7,−6) ⇒ horizontal, $a = 7 − (−2) = 9$, $a^2 = 81$. Focus $(−2 + 2\sqrt{14}, −6)$ ⇒ $c = 2\sqrt{14}$, $c^2 = 56$. So $b^2 = 81 − 56 = 25$. Equation:
$\frac{(x+2)^2}{81} + \frac{(y+6)^2}{25} = 1$

Now compile corrected final answers:

1) $\frac{(x-6)^2}{36} + \frac{(y+8)^2}{16} = 1$
2) $\frac{(x+1)^2}{25} + \frac{(y-6)^2}{81} = 1$
3) $\frac{x^2}{36} + \frac{(y+5)^2}{1} = 1$
4) $\frac{(x+4)^2}{9} + \frac{(y-5)^2}{121} = 1$
5) $\frac{(x-5)^2}{9} + \frac{(y-6)^2}{49} = 1$
6) $\frac{(x-2)^2}{144} + \frac{(y+7)^2}{16} = 1$
7) $\frac{(x-3)^2}{36} + \frac{(y+7)^2}{9} = 1$
8) $\frac{(x-2)^2}{144} + \frac{(y-6)^2}{49} = 1$
9) $\frac{(x-6)^2}{81} + \frac{(y+8)^2}{64} = 1$
10) $\frac{(x+2)^2}{81} + \frac{(y+6)^2}{25} = 1$

All verified.

Final Answer:
1) $\frac{(x-6)^2}{36} + \frac{(y+8)^2}{16} = 1$
2) $\frac{(x+1)^2}{25} + \frac{(y-6)^2}{81} = 1$
3) $\frac{x^2}{36} + \frac{(y+5)^2}{1} = 1$
4) $\frac{(x+4)^2}{9} + \frac{(y-5)^2}{121} = 1$
5) $\frac{(x-5)^2}{9} + \frac{(y-6)^2}{49} = 1$
6) $\frac{(x-2)^2}{144} + \frac{(y+7)^2}{16} = 1$
7) $\frac{(x-3)^2}{36} + \frac{(y+7)^2}{9} = 1$
8) $\frac{(x-2)^2}{144} + \frac{(y-6)^2}{49} = 1$
9) $\frac{(x-6)^2}{81} + \frac{(y+8)^2}{64} = 1$
10) $\frac{(x+2)^2}{81} + \frac{(y+6)^2}{25} = 1$