Question

HELP WILL GIVE BRAINLIEST + 100 POINTS

Answer 1

Answer:

A)  (-4, 0)  and  (36, 0)

B)  Comet H

C)  Focus

Step-by-step explanation:

Part A

Given equation for Comet E

$\dfrac{(x-16)^2}{400}+\dfrac{y^2}{144}=1$

The path of Comet E has been modeled as a horizontal ellipse.

General equation of a horizontal ellipse

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$

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:

$\implies h=16$

$\implies k=0$

$\implies a^2=400 \implies a=20$

$\implies b^2=144 \implies b=12$

Therefore, the vertices are:

$\begin{aligned}\implies (h \pm a, k) & = (16 \pm 20, 0) \\ & = (16-20,0) \:\:\textsf {and }\:(16+20, 0)\\ & = (-4,0) \:\:\textsf {and }\:(36, 0)\end{aligned}$

Part B

Given equation for Comet H:

$\dfrac{(x+13)^2}{144}-\dfrac{y^2}{25}=1$

The path of Comet H has been modeled as a horizontal hyperbola.

General equation of a horizontal hyperbola (opening left and right):

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

where:

• center = (h, k)
• Vertices = (h±a, k)
• Co-vertices = (h, k±b)
• Foci = (h±c, k)  where c²=a²+b²
• $\textsf{asymptotes}: \quad y =k \pm \left(\dfrac{b}{a}\right)(x-h)$
• Transverse axis: y = k
• Conjugate axis: x = h

Comparing the general equation with the given equation:

$\implies h=-13$

$\implies k=0$

$\implies a^2=144 \implies a=12$

$\implies b^2=25 \implies b=5$

Therefore, the vertices are:

$\begin{aligned}\implies (h \pm a, k) & = (-13 \pm 12, 0) \\ & = (-13-12,0) \:\:\textsf {and }\:(-13+12, 0)\\ & = (-25,0) \:\:\textsf {and }\:(-1, 0)\end{aligned}$

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

$a^2=400, \quad b^2=144$

$\begin{aligned}\implies c^2 & =a^2-b^2\\& = 400-144\\& = 256\\\implies c & = \sqrt{256}\\& = 16\end{aligned}$

$\begin{aligned}\implies \sf Foci & = (h \pm c, k)\\ & = (16 \pm 16, 0) \\ & = (16-16,0) \:\:\textsf {and }\:(16+16, 0)\\ & = (0,0) \:\:\textsf {and }\:(32, 0)\end{aligned}$

Foci of Comet H

$a^2=144, \quad b^2=12$

$\begin{aligned}\implies c^2 & =a^2+b^2\\& = 144+25\\& = 169\\\implies c & = \sqrt{169}\\& = 13\end{aligned}$

$\begin{aligned}\implies \sf Foci & = (h \pm c, k)\\ & = (-13 \pm 13, 0) \\ & = (-13-13,0) \:\:\textsf {and }\:(-13+13, 0)\\ & = (-26,0) \:\:\textsf {and }\:(0, 0)\end{aligned}$

Learn more about ellipses here:

brainly.com/question/28152904

Learn more about hyperbolas here:

brainly.com/question/28164074