Condition for a Line to Be a Normal

For the parabola $$y^2=4ax,$$ the equation of the normal at parameter $$t$$ is $$y=-tx+2at+at^3$$

Hence slope of the normal is $$-t$$

Given slope is $$m,$$

therefore

$$t=-m$$

Substituting,

$$y=mx-2am-am^3$$

Thus the line is $$y=mx+c$$

where $$c=-a(2m+m^3)$$

Now this line is tangent to the rectangular hyperbola $$xy=a^2$$

Substitute

$$y=mx+c$$ into $$xy=a^2$$

We get

$$x(mx+c)=a^2$$

$$mx^2+cx-a^2=0$$

For tangency, discriminant must be zero.

Hence,

$$c^2-4(m)(-a^2)=0$$

$$c^2+4a^2m=0$$

Substitute $$c=-a(2m+m^3),$$

$$a^2(2m+m^3)^2+4a^2m=0$$

Divide by $$a^2,$$

$$(2m+m^3)^2+4m=0$$

$$m^2(m^2+2)^2+4m=0$$

$$m\left[m(m^2+2)^2+4\right]=0$$

Since $$m=0$$ does not satisfy tangency for $$xy=a^2,$$

we get

$$m(m^2+2)^2+4=0$$

Expanding,

$$m(m^4+4m^2+4)+4=0$$

$$m^5+4m^3+4m+4=0$$

Hence, the required condition is $$\boxed{m^5+4m^3+4m+4=0}$$