# Hyperbola .Find the points on the hyperbola x^2/9-y^2/4-1=0 that have x=4 .

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### 2 Answers

To determine which points on the hyperbola x^2/9-y^2/4-1=0 that have x=4, we substitute x in the equation of the hyperbola and solve the resulting quadratic equation.

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

=> 16/9 - y^2/4 - 1 = 0

=> 64 - 9y^2 - 36 = 0

=> 9y^2 = 28

=> y^2 = 28/9

y = sqrt (28/9) and -sqrt(28/9)

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

**The points on the hyperbola that have x = 4 are (4, (2/3)*sqrt 7) and (4, (-2/3)*sqrt 7)**

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)}.**