Reflection of a graph about the line $y=2x$.

Question

In the xy-plane, which of the following is the reflection of the graph of $$y=\frac{1+x}{1+x^2}$$ about the line $y=2x$?

a) $x = \frac{1+y}{1+y^2}$ ; b) $x = \frac{-1+y}{1+y^2}$ ; c) $x = -\frac{1+y}{1+y^2}$ ; d) $x = \frac{1-y}{1+y^2}$ ; e) none of the above.

Frankly, I'm not sure if I know enough to get started on this problem. So far, the only thing I could think of was that if you reflect a function over $y=x$ then you simply 'switch' the places of $x$ and $y$, while for reflection over $y=-x$ you switch $x$ and $y$'s and negate both of them. Any help would really be appreciated!

Answer 1

Write $t= 2x$, then we have to find a reflection of $y= {4+2t\over 4+t^2}$ across $y=t$. So replace $y$ and $t$ in given equation and express $y$ in this new equation: $$ t={4+2y\over 4+y^2}$$ So $$ 4t+ty^2=4+2y\implies ty^2-2y+4t-4=0$$ so $$ y= {2\pm 2\sqrt{1-4t^2+4t}\over 2t} = {1\pm \sqrt{-x^2+2x+1}\over 2x}$$ so non of the offered.

Answer 2

You could use a process of elimination. Choice (a) interchanges $x$ and $y$, which you know is reflection in $x=y$, so that’s not it. Choice (c) is choice (a) flipped upside-down, so that’s not it, either. Choice (d) results from the substitutions $x\to-y$ and $y\to x$, which is equivalent to a 90° rotation, so that’s not it, either. The remaining choice is (d) flipped over, which isn’t right, either. You also know that neither of these latter two can be correct because they don’t map the line $y=2x$ to itself, which is what the correct transformation must do. In fact, you can eliminate all of the choices on that basis alone.