Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five

Question

Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1

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Parker 2 weeks 2021-11-20T17:05:44+00:00 2 Answers 0

Answers ( )

    0
    2021-11-20T17:07:02+00:00

    Answer:

    The vertices are (3 , -5) , (-5 , -5)

    The foci are (4 , -5) , (-6 , -5)

    Step-by-step explanation:

    * Lets study the equation of the hyperbola

    – The standard form of the equation of a hyperbola with  

      center (h , k) and transverse axis parallel to the x-axis is

      (x – h)²/a² – (y – k)²/b² = 1

    – The length of the transverse axis is 2 a

    – The coordinates of the vertices are  (h  ±  a  ,  k)

    – The coordinates of the foci are (h ± c , k), where c² = a² + b²

    – The distance between the foci is  2c

    * Now lets solve the problem

    – The equation of the hyperbola is (x + 1)²/16 – (y + 5)²/9 = 1

    * From the equation

    # a² = 16 ⇒ a = ± 4

    # b² = 9 ⇒ b = ± 3

    # h = -1

    # k = -5

    ∵ The vertices are (h + a , k) , (h – a , k)

    ∴ The vertices are (-1 + 4 , -5) , (-1 – 4 , -5)

    * The vertices are (3 , -5) , (-5 , -5)

    ∵ c² = a² + b²

    ∴ c² = 16 + 9 = 25

    ∴ c = ± 5

    ∵ The foci are (h ± c , k)

    ∴ The foci are (-1 + 5 , -5) , (-1 – 5 , -5)

    * The foci are (4 , -5) , (-6 , -5)

    0
    2021-11-20T17:07:17+00:00

    Answer:

    Vertices: (3,-5) (-5,-5)

    Foci: (-6,-5) (4,-5)

    Step-by-step explanation:

    (x+1)^2/16-(y+5)^2/9 =1

    formula: (x-h)^2/a^2 -(y-k)^2/b^2=1

    in this case…

    a^2=16      b^2=9

    h=-1 k=-5

    a=4           b=3

    v=(h+/-a,k)

    v1=(-1+4,-5)=

    v1=(3,-5)

    v2=(-1-4, -5) =

    v2=(-5,-5)

    Foci=(h+/-c,k)

    F1=(h-c,k)

    =(-1-5,-5)

    f1=(-6,-5)

    F2=(h+c,k)

    =(-1+5, -5)

    F2=(4,-5)

    Hope this helps! 🙂

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