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Which hyperbola has one focus in common with the hyperbola x^2/16 - y^2/9 = 1

Which hyperbola has one focus in common with the hyperbola x^2/16 - y^2/9 = 1-example-1
User Inser
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2 Answers

0 votes

Answer:
((y+13)^(2) )/(144) -((x+5)^(2) )/(25) =1

Step-by-step explanation: I got this right on Edmentum.

Which hyperbola has one focus in common with the hyperbola x^2/16 - y^2/9 = 1-example-1
User Anton Hlinisty
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1 vote

Answer:

The same focus is (-5 , 0) ⇒ Answer D

Explanation:

* Lets study the equation of the hyperbola

# The standard form of the equation of a hyperbola with

center (0 , 0) and transverse axis parallel to the x-axis is

x²/a² - y²/b² = 1

- The coordinates of the foci are (± c , 0), where c² = a² + b²

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

# The standard form of the equation of a hyperbola with

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

(y - k)²/a² - (x - h)²/b² = 1

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

* Now lets solve the problem

∵ x²/16 - y²/9 = 1

∴ a² = 16 and b² = 9

∵ c² = a² + b²

∴ c² = 16 + 9 = 25 ⇒ take √ to find the values of c

∴ c = ±√25 = ± 5

∴ The foci are (5 , 0) , (-5 , 0)

# Answer A:

∵ (y - 5)/16 - (x - 13)/9 = 1

∵ (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = 13 and k = 5

∵ a² = 16 and b² = 9

∵ c² = a² + b²

∴ c² = 16 + 9 = 25 ⇒ take √ to find the values of c

∴ c = ±√25 = ± 5

∴ The foci are (13 , 5+5) , (13 , 5-5)

∴ The foci are (13 , 10) , (13 , 0) ⇒ not the same

# Answer B:

∵ (x - 13)²/25 - (y - 5)²/144

∵ (x - h)²/a² - (y - k)²/b² = 1

∵ The foci are (h ± c , k)

∴ h = 13 and k = 5

∵ a² = 25 and b² = 144

∵ c² = a² + b²

∴ c² = 125 + 144 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (13 + 13 , 5) , (13 - 13 , 5)

∴ The foci are (26 , 5) , (0 , 5) ⇒ not the same

# Answer C:

∵ (y - 5)/25 - (x - 13)/144 = 1

∵ (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = 13 and k = 5

∵ a² = 25 and b² = 144

∵ c² = a² + b²

∴ c² = 25 + 144 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (13 , 5+13) , (13 , 5-13)

∴ The foci are (13 , 18) , (13 , -8) ⇒ not the same

# Answer D:

∵ (y + 13)/144 - (x + 5)/25 = 1

∵ (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = -5 and k = -13

∵ a² = 144 and b² = 25

∵ c² = a² + b²

∴ c² = 144 + 25 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (-5 , -13+13) , (-5 , -13-13)

∴ The foci are (-5 , 0) , (-5 , -26) ⇒ one of them the same

* The same focus is (-5 , 0)

User Nat Chouf
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5.7k points