Find the minimum value of $(\alpha-\beta)^2+(\sqrt{2-\alpha^2}-\frac{9}{\beta})^2$
where $ 0<\alpha<\sqrt{2}$ and $\beta>0$
My attempt:
In my view,this minimum value is the shortest distance between circle $x^2+y^2=2$ and rectangular hyberbola $xy=9$.But i dont know how to find.Any hints will help me.
On
the minimum distance between a hyperbola and an ellipse having the same transverse axis is the distance between their vertices. and the vertex of a hyperbola or an ellipse is the point at which the transverse axis and hyperbola or ellipse intersect. here have a special case and the common transverse axis between hyperbola and circle is the line $y=x$
find the intersection of the line y=x with the circle and the hyperbola. it will be (1,1) and (3,3) the the square of the distance between these two points is 8
Your view is right: if you draw the circle and the hyperbola it is apparent that the shortest distance $AB$ is measured along line $y=x$. Point $A$ lies on the circle, so $A(1,1)$, point $B$ is on the hyperbola: $B(3,3)$, so that $AB=2\sqrt2$ and $AB^2=8$.