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(a) Horizontal hyperbola with center ( 0 , 0 ) (b) Vertical hyperbola with center ( 0 , 0 )

Given the equation of a hyperbola in standard form, locate its vertices and foci.

  1. Determine whether the transverse axis lies on the x - or y -axis. Notice that a 2 is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
    1. If the equation has the form x 2 a 2 y 2 b 2 = 1 , then the transverse axis lies on the x -axis. The vertices are located at ( ± a , 0 ) , and the foci are located at ( ± c , 0 ) .
    2. If the equation has the form y 2 a 2 x 2 b 2 = 1 , then the transverse axis lies on the y -axis. The vertices are located at ( 0 , ± a ) , and the foci are located at ( 0, ± c ) .
  2. Solve for a using the equation a = a 2 .
  3. Solve for c using the equation c = a 2 + b 2 .

Locating a hyperbola’s vertices and foci

Identify the vertices and foci of the hyperbola    with equation y 2 49 x 2 32 = 1.

The equation has the form y 2 a 2 x 2 b 2 = 1 , so the transverse axis lies on the y -axis. The hyperbola is centered at the origin, so the vertices serve as the y -intercepts of the graph. To find the vertices, set x = 0 , and solve for y .

1 = y 2 49 x 2 32 1 = y 2 49 0 2 32 1 = y 2 49 y 2 = 49 y = ± 49 = ± 7

The foci are located at ( 0, ± c ) . Solving for c ,

c = a 2 + b 2 = 49 + 32 = 81 = 9

Therefore, the vertices are located at ( 0, ± 7 ) , and the foci are located at ( 0 , 9 ) .

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Identify the vertices and foci of the hyperbola with equation x 2 9 y 2 25 = 1.

Vertices: ( ± 3 , 0 ) ; Foci: ( ± 34 , 0 )

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Writing equations of hyperbolas in standard form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Hyperbolas centered at the origin

Reviewing the standard forms given for hyperbolas centered at ( 0 , 0 ) , we see that the vertices, co-vertices, and foci are related by the equation c 2 = a 2 + b 2 . Note that this equation can also be rewritten as b 2 = c 2 a 2 . This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

Given the vertices and foci of a hyperbola centered at ( 0 , 0 ) , write its equation in standard form.

  1. Determine whether the transverse axis lies on the x - or y -axis.
    1. If the given coordinates of the vertices and foci have the form ( ± a , 0 ) and ( ± c , 0 ) , respectively, then the transverse axis is the x -axis. Use the standard form x 2 a 2 y 2 b 2 = 1.
    2. If the given coordinates of the vertices and foci have the form ( 0, ± a ) and ( 0, ± c ) , respectively, then the transverse axis is the y -axis. Use the standard form y 2 a 2 x 2 b 2 = 1.
  2. Find b 2 using the equation b 2 = c 2 a 2 .
  3. Substitute the values for a 2 and b 2 into the standard form of the equation determined in Step 1.

Questions & Answers

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If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
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SIMRAN Reply
-42m²+60m-18
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-24m+3+3mÁ^2
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-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
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A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
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The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
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Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
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12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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