Question

Find the coordinatesof the point on the curve xy=10 in the first quadrant such that the normal line passes through the orgin.

My work: Okay i believe this is an inverse variation. Therefore, y=10/x? Origin is @ (0,0) and i'm stuck to move on...

Answer 1

since y = 10/x, the slope of the tangent at (x,y) is -10/x^2

so the normal has slope x^2/10.

So, we have a line with slope y/x = (10/x)/(x^2/10) = 100/x^3

so, where does 10/x = 100/x^3?
x = √10

So, at (√10,√10), the slope is 1, and the normal line is just

y = x

Answer 2

Thanks I understand it now :D