There is exactly one value of $x$ for which the distance from $(5,6)$ to $(3x-1,ax+5)$ is $4$. If $a \neq 0,$ what is $a$?
1 answer
We use the distance formula: $$\sqrt{(3x-1-5)^2+(ax+5-6)^2}=4.$$Squaring both sides, we get $$(3x-6)^2+a^2x^2=16.$$Expanding $(3x-6)^2$, we get $$9x^2-36x+36+a^2x^2=16,$$so $(9+a^2)x^2-36x+20=0.$ This quadratic has exactly one root if and only if the discriminant is equal to zero; thus, we want $(36)^2-4(9+a^2)(20)=0.$ Simplifying, we have $1296-720-80a^2=0,$ or $a^2=8.$ Since $a$ is nonzero, we have $a=\boxed{2\sqrt{2}}$.