show that the area of a triangle inscribed in a parabola y^2= 4ax is 1/8 a (y1-y2)(y2-y3)(y3-y1) where y1,y2,y3 are - Maths - Conic Sections

Question

show that the area of a triangle inscribed in a parabola y^2= 4ax
is 1/8 a (y1-y2)(y2-y3)(y3-y1) where y1,y2,y3 are angular points

Rishabh Mittal · Accepted Answer

Solution :
Let the coordinates of the vertices of triangle be x1,y1 , x2,y2 and x3,y3 
As these points will lie on the parabola 
So,y12=4ax1x1=y124aSimilarly ,x2=y224a   and  x3=y324a 
Now,Vertices of the triangle arey124a,y1 , y224a,y2 and y324a,y3So, Area of triangle=12 y124ay11y224ay21y324ay31Taking out 14a common from first column 
=18a y12y11y22y21y32y31Applying R1→R1-R2 , R2→R2-R3=18a y12-y22y1-y20y22-y32y2-y30y32y31=18a y1-y2y1+y2y1-y20y2-y3y2+y3y2-y30y32y31Taking y1-y2 common from R1 and y2-y3 common from R2 =18a y1-y2 y2-y3 y1+y210y2+y310y32y31Expanding along C3=18a y1-y2 y2-y3 11y1+y2-1y2+y3=18a y1-y2 y2-y3 y1-y3=18a y1-y2 y2-y3 y3-y1    sq units

show that the area of a triangle inscribed in a parabola y^2= 4ax is 1/8 a (y1-y2)(y2-y3)(y3-y1) where y1,y2,y3 are angular points.