A line passes through the origin $O$ and cut the parabola $y=-\frac{x^2}{2}+1$ at point B in the first quarter, and also cut the line $y=-x+2$ at point $A$.
Need to find points $A,B$ so that the ration $\frac{AB}{BO}$ will be minimal. Also need to find the equation of that line.
I called my line $y=mx+n$. Then $A(x,-x+2), B(t,-\frac{t^2}{2}+1)$. Then by the distance formula I got
$$f^2=\frac{(-x+2+\frac{t^2}{2}-1)^2+(x-t)^2}{(\frac{t^2}{2}-1)^2+t^2}$$
But I have too many letters here, don't know how to get out of it.
Thanks.
let the line through the origin be $y=mx$ and let $m=\tan\theta$. From the data given, at B the $x$ component is $-m+\sqrt{m^2+2}=p$ and at A the $x$ component is $\frac{2}{m+1}=q$.
The ratio $$\frac{AB}{OB}=\frac{AB\cos\theta}{OB\cos\theta}$$, so it is sufficient to consider the ratio of the $x$ components $\frac{q-p}{p}=\frac qp-1$
This will be minimum when the quantity $$R=\frac pq=\frac 12(m+1)(-m+\sqrt{m^2+2})$$ is maximum.
Considering $\frac{dR}{dm}=0$, and a bit of algebra which reduces very easily, we end up with $$m=\frac 12$$