Oxford Entrance Exam Problem 5

If you're allowed to use the options, you can rule out all but one pretty quickly. First, let's say the reflected point is in the form $\left(f(m), g(m)\right)$ . Right away, we know that $\begin{cases} \lim _{ m\rightarrow \infty }{ f(m) } =-1 \\ \lim _{ m\rightarrow \infty }{ g(m) } =0 \end{cases}$ This leaves only two possibilities for $g(m)$ : $\frac { 2m }{ 1+{ m }^{ 2 } } , \frac { m }{ { m }^{ 2 }-1 }$ . The second option's accompanying $f(m)$ has a limit of $1$ and not $-1$ , leaving one option.

If not, there's a simple geometric approach:

We have that $a+{m}^{2}a=1$ so $a=\frac { 1 }{ { m }^{ 2 }+1 }$ . We now have that the $y$ coordinate is $\frac { 2m }{ 1+{ m }^{ 2 } }$ .

The $x$ is a little bit trickier, but we can see that it's just $a-{m}^{2}a$ which gives a final result for the $x$ as $\frac { 1-{ m }^{ 2 } }{ { m }^{ 2 }+1 }$ .

In the above graph drawn, we are constructing a perpendicular line from $(1,0)$ to the line $y = mx$

Using slope-point formula, the equation of the constructed line would be

$y +\frac{x}{m} = \frac{1}{m}$

Equating the above line and $y =mx$ , we get the point of intersection as $(\frac{1}{1+m^2} , \frac{m}{1+m^2})$

The point of intersection is the mid point of the line joining the reflection and the object.

Let the reflection be $(\alpha,\beta)$

Using mid-point formula, we get $(\alpha,\beta) =(\frac{1-m^2}{1+m^2}, \frac{2m}{1+m^2})$