Question #60f1a

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Answer 1 · 2016-11-03T17:46:36

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

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)#