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
ab approve bhi krde bhai, itne answer dediye :)
						

							
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