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Then the equation of the tangent at $A=(1,1)$ is of the form $$y-1=m(x-1),$$or $y=mx-m+1.$ Substituting into $y=x^2,$ we get $$mx-m+1=x^2.$$Then $x^2-mx+m-1=0.$ Since we have a tangent, this quadratic should have a double root. And since the $x$-coordinate of $A$ is $1,$ the double root is $x=1.$ Hence, this quadratic is identical to $(x-1{)}^2=x^2-2x+1,$ which means $m=2.$

Then the slope of the normal is $-\frac{1}{2},$ so the equation of the normal is $$y-1=-\frac{1}{2}(x-1).$$We want the intersection of the normal with $y=x^2,$ so we set $y=x^2$: $$x^2-1=-\frac{1}{2}(x-1).$$We can factor the left-hand side: $$(x-1)(x+1)=-\frac{1}{2}(x-1).$$The solution $x=1$ corresponds to the point $A.$ Otherwise, $x\ne 1,$ so we can divide both sides by $x-1$: $$x+1=-\frac{1}{2}.$$Hence, $x=-\frac{3}{2},$ so $B=\boxed{\left(-\frac{3}{2},\frac{9}{4}\right)}.$