jmc

Let $A=(a,a^2).$ Then the equation of the tangent at $A$ is of the form $$y-a^2=m(x-a).$$Setting $y=x^2,$ we get $x^2-a^2=m(x-a),$ or $x^2-mx+ma-a^2=0.$ Since we have a tangent, this quadratic will have a double root of $x=a$; in other words, this quadratic is identical to $x^2-2ax+a^2=0.$ Hence, $m=2a.$

Therefore, the equation of the tangent at $A$ is $$y-a^2=2a(x-a).$$Similarly, the equation of the tangent at $B$ is $$y-b^2=2b(x-b).$$To find the point of intersection $P,$ we set the value of $y$ equal to each other. This gives us $$2a(x-a)+a^2=2b(x-b)+b^2.$$Then $2ax-a^2=2bx-b^2,$ so $$(2a-2b)x=a^2-b^2=(a-b)(a+b).$$Since $a\ne b,$ we can divide both sides by $2a-2b,$ to get $$x=\frac{a+b}{2}.$$Then $$\begin{array}{rl}y& =2a(x-a)+a^2\\ & =2a\left(\frac{a+b}{2}-a\right)+a^2\\ & =a^2+ab-2a^2+a^2\\ & =ab.\end{array}$$Note that the two tangents are perpendicular, so the product of their slopes is $-1.$ This gives us $(2a)(2b)=-1.$ Hence, the $y$-coordinate of $P$ is always $ab=\boxed{-\frac{1}{4}}.$ This means that the intersection point $P$ always lies on the directrix $y=-\frac{1}{4}.$