I have attached the image of the problem.Please help me with this problem giving detailed step by step answer.

we see that if we consider 2 graphs y= f(x)= sqrt(2 – x^2) and y= g(x)= 9/x, then the distance b/w a point (x1, f(x1)) on y= f(x) and another point (x2, g(x2)) on y= g(x) is given by

D^2= (x1 – x2)^2 + (f(x1) – g(x2))^2

so, to minimise this, we need to draw graphs of these fns.

obviously y= f(x) is a circle given by x^2+y^2= 2, with centre at origin and radius root2 while y= g(x) is a rectangular hyperbola xy= 9.

after u draw the figures, u ll see that min distance will occur along the common normal y= x, which intersects y= f(x) at (1, 1) and y= g(x) at (3, 3).

so, D^2|min = (3 – 1)^2 + (3 – 1)^2

= 8

@Aditya gupta (1725 points) !!! i posted this question because i did not understand how y=x is the common normal.I thought i can understand how y-x is the common normal from askiitians forum.But still could not understand  how y=x is the common normal.

note that any normal to the circle has to pass through its centre (0,0).

so assume the normal is y= mx with slope m.

now, it intersects xy= 9 at mx^2= 9 or x= 3/sqrt(m) as x is positive.

now, slope of tangent at this point on hyperbola is dy/dx= – 9/x^2= – m

so, since tangent and normal are perpendicular, m*( – m) = – 1

or m= ± 1

but since x= 3/sqrt(m), m cant be – ve.

so, m= 1

or common normal is y= 1*x

or y=x