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Tangents are drawn to the parabola at three distinct points. Prove thatthese tangent lines always make a triangle and that the locus of theorthocentre of the triangle is the directrix of the parabola. |
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Answer» 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. |
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