Given the equation 2x^2 - 50y^2 = 50 determine: a) The Center C b) The 2 Vertices c) The slopes of the asymptotes

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

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Given `2x^2-50y^2=50` :

Divide through by 50 to get:

`x^2/25-y^2/1=1`

The standard form for a hyperbola is `(x-h)^2/a^2-(y-k)^2/b^2=1` with center (h,k) and horizontal transverse axis.

Since h=k=0 the center is at the origin.

The vertices are `(+-a,0)` ; since a=5 we have (5,0) and (-5,0)

With center (0,0) you can form a rectangle whose center is at the origin and whose side lengths are 2a and 2b -- since the transverse axis is horizontal the horizontal width is 10 and the vertical height is 2. The corners of the rectangle are (5,1),(-5,1),(-5,-1) and (5,-1).

The asymptotes for the hyperbola are the diagonals of this rectangle.The slopes of the diagonals are `-1/5` and `1/5` .

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C is at (0,0)

The vertices are at (5,0) and (-5,0)

The slopes of the asymptotes are `m=+-1/5`

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The graph:

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