How do you write a standard form equation for the hyperbola with asymptotes are at y= 1/2 x and y= -1/2x. The hyperbolas vertices are at (-3,0) and (3,0)?

1 Answer
Jan 20, 2017

Answer:

#x^2/(3/2)^2-y^2/(3/2)^2=1#. See the asymptotes-inclusive graph of the rectangular hyperbola (RH) in the Socratic graph.

Explanation:

If A1=9 and A2 =0 are the equations to the asymptotes, the quation

to the family of hyperbolas with thesr aymptotes is

#A1xxA2= c#. c is called the parameter.

Here, it is

(y-x/2)(y+x/2)= c

The parameter value for the member of this family that has the

vertices A( 3, 0) and A'(-3, 0) is given by

#(0-3/2)(0+3/2)=-9/4=c.#

Now, the equation is in the standard form

#x^2/(3/2)^2-y^2/(3/2)^2=1#, revealing that it is rectangular.

See the asymptotes-inclusive RH in the graph.

graph{(x^2-y^2-9/4)(x^2-y^2/4)=0 [-10, 10, -5, 5]}