Tangents are drawn to the parabola at three distinct points. Prove that these tangent lines always make a triangle and that the locus of the orthocentre of the triangle is the directrix of the parabola.

Explanation

Let the three point on the parabola be `P(at_(1)^(2),2at_(1)),Q(at_(2)^(2),2at_(2)),R(at_(3)^(2),2at_(3))`.
Points of intersection of tangents drawn at these point are
`A(at_(2)t_(3),a(t_(2)+t_(3)),B(at_(1)t_(3),a(t_(1)+t_(3)))andC(at_(1)t_(2),a(t_(1)+t_(2)))`.
To find the orthocentre of the triangle ABC, we need to find equation of altitudes of the triangle.
Slope of BC `=(a(t_(2)-t_(3)))/(at_(1)(t_(2)-t_(3)))=(1)/(t_(1))`
Thus, slope of altitude through vertex A is `-t_(1)`.
Equation of altitude through A is
`y-a(t_(2)+t_(3))=-t_(1)(x-at_(2)t_(3))`
`or" "y-a(t_(2)+t_(3))=-t_(1)x+at_(1)t_(2)t_(3)` (1)
Similarly, equation of altitude through B is
`y-a(t_(1)+t_(3))=-t_(2)xat_(1)t_(2)t_(3)` (2)
Subtracting (2) from (1), we get `a(t_(1)-t_(2))=(t_(2)-t_(1))xorx=-a`.
Thus , orthocentre lies on the directrix.