# Draw the set of points (x; y) in the plane which satisfy y^2-2xy- 8x^2 = 0 . Is this set of points the graph of a function?

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### 1 Answer

You should consider `y!=0` , hence, you may divide by `y^2` such that:

`1 - 2(x/y) - 8(x/y)^2 = 0`

You need to come up with the following substitution such that:

`x/y = t`

Changing the variable yields:

`1 - 2t - 8t^2 = 0 => -8t^2 - 2t + 1 = 0`

`8t^2 + 2t - 1 = 0`

You need to use the quadratic formula such that:

`t_(1,2) = (-2+-sqrt(4+32))/16 => t_(1,2) = (-2+-sqrt36)/16`

`t_(1,2) = (-2+-6)/16 => t_1 = 4/16 = 1/4`

`t_2 = -8/16 = -1/2`

You should substitute back `-8/16` and `1/4` for `x/y` such that:

`-8/16 = x/y => 16x = -8y => 2x = -y => x = -y/2`

`1/4 = x/y => 4x = y`

**Hence, evaluating the points that satisfy the given equation yields that they need to satisfy the following equations`y = -2x` and `y = 4x` .**