jmc

Let $(x,y)$ be a point on the hyperbola $x^2+8xy+7y^2=225.$ Effectively, we want to minimize $x^2+y^2.$ Let $k=x^2+y^2.$ Multiplying this with the equation $x^2+8xy+7y^2=225,$ we get $$k{x}^2+8kxy+7k{y}^2=225x^2+225y^2,$$so $$(k-225)x^2+8kxy+(7k-225)y^2=0.$$For the curves $x^2+8xy+7y^2=225$ and $x^2+y^2=k$ to intersect, we want this quadratic to have a real root, which means its discriminant is nonnegative: $$(8ky{)}^2-4(k-225)(7k-225)y^2\ge 0.$$This simplifies to $y^2(36k^2+7200k-202500)\ge 0,$ which factors as $$y^2(k-25)(k+225)\ge 0.$$If $y=0,$ then $x^2=225,$ which leads to $x=\pm 15.$ The distance from the origin to $P$ is then 15. Otherwise, we must have $k=x^2+y^2\ge 25.$

If $k=25,$ then $-200x^2+200xy-50y^2=-50(2x-y{)}^2=0,$ so $y=2x.$ Substituting into $x^2+8xy+7y^2=225,$ we get $45x^2=225,$ so $x^2=5,$ which implies $x=\pm \sqrt{5}.$ Thus, equality occurs when $(x,y)=(\sqrt{5},2\sqrt{5})$ or $(-\sqrt{5},-2\sqrt{5}),$ and the minimum distance from the origin to $P$ is $\boxed{5}.$