Maths-
General
Easy

Question

Area of triangle formed by the tangents at three points t1, t2 and t3 of the parabola y2 = 4ax is-

  1. fraction numerator a over denominator 2 end fraction (t1 – t2) (t2– t3) (t3– t1)    
  2. a2 (t1 – t2) (t2– t3) (t3– t1)    
  3. fraction numerator a to the power of 2 end exponent over denominator 2 end fraction (t1 + t2) (t2+ t3) (t3+ t1)    
  4. fraction numerator a to the power of 2 end exponent over denominator 2 end fraction (t1 – t2) (t2– t3) (t3– t1)    

hintHint:

find the vertices of the triangle and apply the determinant rule to find the area under the triangle.

The correct answer is: fraction numerator a to the power of 2 end exponent over denominator 2 end fraction (t1 – t2) (t2– t3) (t3– t1)


    a2/2(t1-t2)(t2-t3)(t3-t1)
    The parameters for the 3 points are given to be: t1, t2 and t3.
    The parametric points are: (at12,2at1), (at22,2at2) and (at32,2at3)
    We know that area under three points of a triangle are given by the determinant
    A=1/2| a1   b1   1|
    |a2   b2   1|
    |a3   b3   1|
    Where (a1,b1),(a2,b2) and (a3,b3) are the vertices of the triangle.
    This gives us
    a2/2(t1-t2)(t2-t3)(t3-t1)

    area under a triangle is given by the determinant of the vertex points of the triangle.

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