Question #60f1a

1 Answer
Nov 3, 2016

P_(1,2)=(a+-a/bsqrt(a^2+b^2);0)

Explanation:

Let P=(x_0;0) on the x axis
x/a+y/b=1\ \ \ is equivalent to
bx+ay-ab = 0
The distance of the generic P to this line is

abs(bx_0+a0-ab)/sqrt(a^2+b^2)\ \ \ and must be =a

so

abs(bx_0-ab)=asqrt(a^2+b^2)

(bx_0-ab)^2=a^2(a^2+b^2)

b^2x_0^2+a^2b^2-2ab^2x_0=a^2(a^2+b^2)

b^2x_0^2-2ab^2x_0 +a^2b^2-a^4-a^2b^2 =0

x_(1,2)=(ab^2+-sqrt(a^2b^4+a^4b^2))/b^2=a+-a/bsqrt(a^2+b^2)