Respuesta :
Answer:
The standard form of the equation of a hyperbola is [tex]\frac{x^{2}}{16}-\frac{(y-5)^{2}}{20}=1[/tex]
Step-by-step explanation:
The standard form of the equation of a hyperbola with center (h , k) is [tex]\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1[/tex] , where
- The coordinates of the vertices are (h ± a , k)
- The coordinates of the foci are (h ± c , k), where c² = a² + b²
∵ The vertices of the hyperbola are (-4 , 5) and (4 , 5)
∴ k = 5
∴ h + a = 4 ⇒ (1)
∴ h - a = -4 ⇒ (2)
- Add (1) and (2)
∴ 2h = 0
- Divide both sides by 2
∴ h = 0
- Substitute the value of h in (1) or (2) to find a
∵ 0 + a = 4
∴ a = 4
∵ The foci of the hyperbola are (-6 , 5) and (6 , 5)
∴ k = 5
∴ h + c = 6 ⇒ (1)
∴ h - c = -6 ⇒ (2)
- Add (1) and (2)
∴ 2h = 0
- Divide both sides by 2
∴ h = 0
- Substitute the value of h in (1) or (2) to find c
∵ 0 + c = 6
∴ c = 6
Let us use the rule c² = a² + b²
∵ (6)² = (4)² + b²
∴ 36 = 16 + b²
- Subtract 16 from both sides
∴ 20 = b²
∵ The standard form of the equation of a hyperbola is [tex]\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1[/tex]
- Substitute the values of h, k, a and b in the equation
∴ [tex]\frac{(x-0)^{2}}{16}-\frac{(y-5)^{2}}{20}=1[/tex]
∴ [tex]\frac{x^{2}}{16}-\frac{(y-5)^{2}}{20}=1[/tex]
The standard form of the equation of a hyperbola is [tex]\frac{x^{2}}{16}-\frac{(y-5)^{2}}{20}=1[/tex]