What are the foci of the hyperbola represented by the equation (x^(2))/(16)-(y^(2))/(25)=1? Choose 1 answer: (A) (3,0) and (-3,0) (B) (sqrt41,0) and (-sqrt41,0) (c) (0,3) and (0,-3) (D) (0,sqrt41) and (0,-sqrt41)

Equation of the hyperbola: The given equation is of a hyperbola in the standard form (x^2)/(a^2) - (y^2)/(b^2) = 1, where a^2 is under the x^2 term and b^2 is under the y^2 term. For a hyperbola centered at the origin with a horizontal transverse axis, the foci are located at (±c, 0), where c is found using the equation c^2 = a^2 + b^2.Identifying a^2 and b^2: Identify the values of a^2 and b^2 from the given equation. Here, a^2 = 16 and b^2 = 25.Calculating c^2: Calculate the value of c^2 using the equation c^2 = a^2 + b^2. Substituting the identified values gives us c^2 = 16 + 25.Finding the value of c: Perform the addition to find c^2. So, c^2 = 41.Locating the foci: Find the value of c by taking the square root of c^2. Therefore, c = sqrt(41).The foci of the hyperbola are at (±c, 0). Substituting the value of c, we get the foci at (sqrt(41), 0) and (-sqrt(41), 0).

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What are the foci of the hyperbola represented by the equation
(x^(2))/(16)-(y^(2))/(25)=1?
Choose 1 answer:
(A)
(3,0) and
(-3,0)
(B)
(sqrt41,0) and
(-sqrt41,0)
(c)
(0,3) and
(0,-3)
(D)
(0,sqrt41) and
(0,-sqrt41)

What are the foci of the hyperbola represented by the equation $\frac{x^{2}}{16}-\frac{y^{2}}{25}=1 ?$ $\newline$ Choose $1$ answer: $\newline$ (A) $(3,0)$ and $(-3,0)$ $\newline$ (B) $(\sqrt{41}, 0)$ and $(-\sqrt{41}, 0)$ $\newline$ (C) $(0,3)$ and $(0,-3)$ $\newline$ (D) $(0, \sqrt{41})$ and $(0,-\sqrt{41})$

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Q. What are the foci of the hyperbola represented by the equation

\frac{x^{2}}{16}-\frac{y^{2}}{25}=1 ?

\newline

Choose

1

answer:

\newline

(A)

(3,0)

and

(-3,0)

\newline

(B)

(\sqrt{41}, 0)

and

(-\sqrt{41}, 0)

\newline

(C)

(0,3)

and

(0,-3)

\newline

(D)

(0, \sqrt{41})

and

(0,-\sqrt{41})

Equation of the hyperbola: The given equation is of a hyperbola in the standard form $(\frac{x^2}{a^2}) - (\frac{y^2}{b^2}) = 1$ , where $a^2$ is under the $x^2$ term and $b^2$ is under the $y^2$ term. For a hyperbola centered at the origin with a horizontal transverse axis, the foci are located at $(\pm c, 0)$ , where $c$ is found using the equation $c^2 = a^2 + b^2$ .
Identifying $a^2$ and $b^2$ : Identify the values of $a^2$ and $b^2$ from the given equation. Here, $a^2 = 16$ and $b^2 = 25$ .
Calculating $c^2$ : Calculate the value of $c^2$ using the equation $c^2 = a^2 + b^2$ . Substituting the identified values gives us $c^2 = 16 + 25$ .
Finding the value of c: Perform the addition to find $c^2$ . So, $c^2 = 41$ .
Locating the foci: Find the value of $c$ by taking the square root of $c^2$ . Therefore, $c = \sqrt{41}$ .
Locating the foci: Find the value of $c$ by taking the square root of $c^2$ . Therefore, $c = \sqrt{41}$ .The foci of the hyperbola are at $(\pm c, 0)$ . Substituting the value of $c$ , we get the foci at $(\sqrt{41}, 0)$ and $(-\sqrt{41}, 0)$ .