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Olin [163]
3 years ago
10

What are the foci of the ellipse 8x^2+9y^2=96

Mathematics
1 answer:
tino4ka555 [31]3 years ago
8 0

Answer:

Step-by-step explanation:

Let's first get this into standard form of an ellipse.  That means that we divide everything by 96 so we have a 1 on the right side of the equals sign:

\frac{8x^2}{96}+\frac{9y^2}{96}=1

Doing the division simplifies this down to

\frac{x^2}{12}+\frac{y^2}{\frac{32}{3} }=1

We know from the numerators that this is an ellipse centered about the origin (its center is (0, 0).

To find the coordinates of the foci, use

c^2=a^2-b^2

For an ellipse, a is always larger than b, and the larger number is always under the x-term.  So we have a horizontally oriented ellipse.  Our a = 12 and b = 32/3.  Filling in our equation:

c^2=12^2-(\frac{32}{3})^2

and

c^2=144-\frac{1024}{9} and

c^2=\frac{272}{9}

Taking the square root of both sides and simplifying in the process gives you that

c=+/-\frac{4\sqrt{17} }{3}

So the coordinates of the foci are

(-\frac{4\sqrt{17} }{3},0)(\frac{4\sqrt{17} }{3},0)

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Answer:

24.03 units (nearest hundredth)

Step-by-step explanation:

The distance between B and A is:  AB = AH + HB

We have been given AH, so we just need to find the measure of HB.

First, find the angle AOH using tan trig ratio:

\sf \tan(\theta)=\dfrac{O}{A}

where:

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  • O is the side opposite the angle
  • A is the side adjacent the angle

Given:

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\implies \sf \tan(\angle AOH)=\dfrac{8}{19.8}

\implies \sf \angle AOH = 22.00069835^{\circ}

   ∠BOA = ∠BOH + ∠AOH

⇒ ∠BOH = ∠BOA - ∠AOH

⇒ ∠BOH = 61° - 22.00069835°

               = 38.99930165°

Now we can find HB by again using the tan trig ratio:

Given:

  • \theta = ∠BOH = 38.99930165°
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Substituting given values:

\implies \sf \tan(38.99930165^{\circ})=\dfrac{HB}{19.80}

\implies \sf HB=19.80 \tan(38.99930165^{\circ})

\implies \sf HB=16.03332427

Therefore:

   AB = AH + BH

⇒ AB = 8 + 16.03332427

         = 24.03 units (nearest hundredth)

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