Question 750315
the problem states

Q/. Point (a,b) lies in the third quadrant on the graph of the equation y = 1/x. Find a and b given that the distance from point (a,b) to the origin is 
(sqrt(257)/4)

note that a point in the third quadrant has a negative x and negative y.  Also
y = 1/x  has components in the first and third quadrants

The point (a,b) lies on the line y = 1/x in the third quadrant, we know

X^2 + (1/x)^2 = 257 / 16

note that 16 is 4^2

so if x = -4 then y = - 1/4

and 16 + (1/16) = (16^2 + 1) / 16 = 257 / 16