Question

find two straight lines that are tangent to y=x^{2}/(x-1) and pass through the point (2,0)

Answer 1

find the slope of the line that is tangent to y=x^2/(x-1) at (x,y) by differentiating the function,
$$\frac{dy}{dx}=\frac{(x-1)(2x)-x^2}{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}=m$$

if the point that the lines touch is $$(x_0,y_0)$$ $$m=\frac{x_0(x_0-2)}{(x_0-1)^2}$$
also we know that,
$$\frac{y_0-0}{x_0-2}=m\\
\therefore \frac{y_0-0}{x_0-2}=\frac{x_0(x_0-2)}{(x_0-1)^2}\\
y_0=\frac{x_0(x_0-2)^2}{(x_0-1)^2}$$
since (x_0,y_0) are on the curve,
$$y_0=\frac{x_0^2}{(x_0-1)}\\
\therefore \frac{x_0(x_0-2)^2}{(x_0-1)^2}=\frac{x_0^2}{(x_0-1)}\\
x_0(x_0-2)^2=x_0^2(x_0-1)\quad(x_0\not=1)\\
x_0[(x_0-2)^2-x_0(x_0-1)]=0\\
x_0(-3x_0+4)=0\\
x_0=0\;or\;x_0=\frac 43
$$
tangent points are,
(0,0) and (4/3,16/3)

since you know two points of each lines you can find the tangent lines

Answer 2

thank you so much. I'm very pleased to your patience.

Answer 3

Youre welcome :)