(a) find dy/dx if (x-2y)^2=y

Question

(b) Find all the ponts (x,y) where the tangent line to the curve y/x-x/y=5 is either horizontal or vertical

Answer 1

(a) 2(x-2y)(1-2dy/dx)=dy/dx.

2x-4xdy/dx-4y+8ydy/dx=dy/dx,

(dy/dx)(-4x+8y-1)=4y-2x,

dy/dx=(4y-2x)/(8y-4x-1).

(b) This can be written y²-x²=5xy.

2ydy/dx-2x=5xdy/dx+5y,

(2y-5x)dy/dx=2x+5y, dy/dx=(2x+5y)/(2y-5x).

When x=0, y=0, so (0,0) is a point on the graph of this equation. At this point dy/dx is undefined.

y²-5xy-x²=0, so using the quadratic formula:

y=(5x±√(25x²+4x²)/2=(5x±x√29)/2=(5±√29)x, which represents two linear graphs:

y=(5+√29)x/2 and y=(5-√29)x/2. The slopes (gradients) of these two components are perpendicular because (5+√29)(5-√29)/4=-4/4=-1. Therefore there are two slopes: 5.1926 and -0.1926 at right angles. The graphed equation looks like the x and y axes have been tilted clockwise through an angle of about 10.9°. The gradient is therefore never 0 (horizontal) or infinity (vertical) because the inclination of the lines is constant at 10.9° clockwise to the axes.