Answer:
A) (-4, 0) and (36, 0)
B) Comet H
C) Focus
Step-by-step explanation:
Part A
Given equation for Comet E
[tex]\dfrac{(x-16)^2}{400}+\dfrac{y^2}{144}=1[/tex]
The path of Comet E has been modeled as a horizontal ellipse.
General equation of a horizontal ellipse
[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex]
where:
- center = (h, k)
- Vertices = (h±a, k)
- Co-vertices = (h, k±b)
- Foci = (h±c, k) where c²=a²-b²
- 2a = Major Axis: longest diameter of an ellipse
- 2b = Minor Axis: shortest diameter of an ellipse
- a = Major radius: one half of the major axis
- b = Minor radius: one half of the minor axis
Comparing the general equation with the given equation:
[tex]\implies h=16[/tex]
[tex]\implies k=0[/tex]
[tex]\implies a^2=400 \implies a=20[/tex]
[tex]\implies b^2=144 \implies b=12[/tex]
Therefore, the vertices are:
[tex]\begin{aligned}\implies (h \pm a, k) & = (16 \pm 20, 0) \\ & = (16-20,0) \:\:\textsf {and }\:(16+20, 0)\\ & = (-4,0) \:\:\textsf {and }\:(36, 0)\end{aligned}[/tex]
Part B
Given equation for Comet H:
[tex]\dfrac{(x+13)^2}{144}-\dfrac{y^2}{25}=1[/tex]
The path of Comet H has been modeled as a horizontal hyperbola.
General equation of a horizontal hyperbola (opening left and right):
[tex]\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1[/tex]
where:
- center = (h, k)
- Vertices = (h±a, k)
- Co-vertices = (h, k±b)
- Foci = (h±c, k) where c²=a²+b²
- [tex]\textsf{asymptotes}: \quad y =k \pm \left(\dfrac{b}{a}\right)(x-h)[/tex]
- Transverse axis: y = k
- Conjugate axis: x = h
Comparing the general equation with the given equation:
[tex]\implies h=-13[/tex]
[tex]\implies k=0[/tex]
[tex]\implies a^2=144 \implies a=12[/tex]
[tex]\implies b^2=25 \implies b=5[/tex]
Therefore, the vertices are:
[tex]\begin{aligned}\implies (h \pm a, k) & = (-13 \pm 12, 0) \\ & = (-13-12,0) \:\:\textsf {and }\:(-13+12, 0)\\ & = (-25,0) \:\:\textsf {and }\:(-1, 0)\end{aligned}[/tex]
As the sun is located at the origin (0, 0) the comet that travels closer to the sun is the comet whose vertex is closest to (0, 0).
Therefore, Comet H travels closer to the sun, since one of its vertex (-1, 0) is the closest to (0, 0).
Part C
The sun represents one of the foci of both Comet E and Comet H.
Foci of Comet E
[tex]a^2=400, \quad b^2=144[/tex]
[tex]\begin{aligned}\implies c^2 & =a^2-b^2\\& = 400-144\\& = 256\\\implies c & = \sqrt{256}\\& = 16\end{aligned}[/tex]
[tex]\begin{aligned}\implies \sf Foci & = (h \pm c, k)\\ & = (16 \pm 16, 0) \\ & = (16-16,0) \:\:\textsf {and }\:(16+16, 0)\\ & = (0,0) \:\:\textsf {and }\:(32, 0)\end{aligned}[/tex]
Foci of Comet H
[tex]a^2=144, \quad b^2=12[/tex]
[tex]\begin{aligned}\implies c^2 & =a^2+b^2\\& = 144+25\\& = 169\\\implies c & = \sqrt{169}\\& = 13\end{aligned}[/tex]
[tex]\begin{aligned}\implies \sf Foci & = (h \pm c, k)\\ & = (-13 \pm 13, 0) \\ & = (-13-13,0) \:\:\textsf {and }\:(-13+13, 0)\\ & = (-26,0) \:\:\textsf {and }\:(0, 0)\end{aligned}[/tex]
Learn more about ellipses here:
https://brainly.com/question/28152904
Learn more about hyperbolas here:
https://brainly.com/question/28164074
