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Determine the points on the hyperbola x^2/9 - y^2/4 - 1 = 0 that have x=4

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chemstruct | Student, Grade 11 | (Level 2) eNoter

Posted December 22, 2010 at 11:30 PM via web

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Determine the points on the hyperbola x^2/9 - y^2/4 - 1 = 0 that have x=4

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted December 22, 2010 at 11:35 PM (Answer #1)

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We'll note the points located on the hyperbola as M(4 , t).

A point is located on a hyperbola if and only if it's coordinates, substituted in the equation of the hyperbole, they verify it.

4^2/9 - t^2/4 = 1

16/9 - t^2/4 = 1

We'll multipy by 36 both sides:

16*4 - t^2*9 = 36

64 - 9t^2 - 36 = 0

9t^2 = 28

t^2 = 28/9

t1 = +(2 sqrt 7)/3

t2 = - (2 sqrt 7)/3

The points located on the hyperbola are: {(4 ; - (2 sqrt 7)/3) ; (4 ; +(2 sqrt 7)/3)}.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted December 22, 2010 at 11:35 PM (Answer #2)

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Given the hyperbola :

x^2 / 9  - y^2 /4  = 1

To find the points of the hyperbola that have x= 4, we will substitute with x= 4 and determine y values.

==> 4^2 / 9 - y^2 /4 = 1

==> 16/9  - y^2 /4 = 1

==> y^2/4 = 16/9 - 1

==> y^2 /4 = 7/9

Now we will multiply by 4.

==> y^2 = 7*4/9 = 28/9

==> y^2 = 28/9

==> y= +- sqrt28/ 3

          = +-2sqrt7 / 3

==> There are two points on the hyperbola that has x=4.

=> ( 4, 2sqrt7 /3 )   and  ( 4, -2sqrt7 / 3)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 22, 2010 at 11:37 PM (Answer #3)

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For the hyperbola x^2/9 - y^2/4 - 1 = 0, when x = 4 , we have

x^2/9 - y^2/4 - 1 = 0

=> 4^2 / 9 - y^2 / 4 - 1 = 0

=> y^2 / 4 = 4^2 / 9  - 1

=> y^2 / 4 = 16 / 9  - 9/9

=> y^2 / 4 = 7/9

=> y^2 = 7*4/9

=> y = (2/3)*sqrt 7 and y = -(2/3)sqrt 7

So the points on the hyperbola that have x = 4 are

(4 , (2/3)*sqrt 7) and ( 4 , -(2/3)sqrt 7)

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neela | High School Teacher | (Level 3) Valedictorian

Posted December 22, 2010 at 11:48 PM (Answer #4)

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To determine the points on the hyperbola points for which x= 4.

So we put x= 4 in the given equation x^2/9 - y^2/4 - 1 = 0 and solve for y.

=> 4^2/9 -y^2/4 -1 = 0.

We multiply by 36 :

=> 16*4 - 9y^2 - 36 = 0.

=> 64 -36 = 9y^2.

=>28 = 9y^2.

=> 9y^2 = 28.

=> y^2 = 28/9.

=> y = sqrt(28/9) = (2sqrt7)/3, Or y = -(2sqrt7)/3.

Therefore the points on the hyperbola are: (4, (2sqr7/)3) and (4, - (2sqrt7)/3).

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