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Question

Prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.

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Solution

Equation of the parabola is y2=4ax
Let A(at21,2at1),B(at22,2at2) and C(at23,2at3) be the three given points on the given parabola.
Equation of the tangent to the parabola at A(at21,2at1) is
yt1=x+at21 ........(1)
Equation of tangent to the parabola at B(at22,2at2) is
yt2=x+at22 ........(2)
and C(at23,2at3) is
yt3=x+at23 ........(3)
Suppose the tangents at A,B and C intersect at P,Q and R respectively.
Solving (1) and (2) we get
x=at1t2 and y=a(t1+t2)
Co-ordinate of P=(at1t2,a(t1+t2))
Solving (2) and (3) we get
x=at2t3 and y=a(t2+t3)
Co-ordinate of Q=(at2t3,a(t2+t3))
Solving (3) and (1) we get
x=at3t1 and y=a(t3+t1)
Co-ordinate of Q=(at3t1,a(t3+t1))
Area ABC=12at21(2at12at2)+at22(2at32at1)+at23(2at12at2)
This expression on simplification gives
Area of ABC=12a2|(t1t2)(t2t3)(t3t1)|
Area of PQR=12|at1t2[a(t2+t3)a(t1+t3)]+at2t3[a(t1+t3)a(t1+t2)]+at3t1[a(t1+t2)a(t2+t3)]|
This expression on simplification gives
Area of PQR=12×a22|(t1t2)(t2t3)(t3t1)|
Area(ABC)Area(PQR)=a22|(t1t2)(t2t3)(t3t1)|a24|(t1t2)(t2t3)(t3t1)|=2
Area of (ABC)=2× Area of (PQR)


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