If `y_1, y_2, y_3` be the ordinates of a vertices of the triangle inscribed in a parabola `y^3=4a x ,` then show that the area of the triangle is `1/(8a)|(y_1-y_2)(y_2-y_3)(y_3-y_1)|dot`

Options

A. `(1)/(2a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`
B. `(1)/(4a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`
C. `(1)/(8a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`
D. none of these

Answer

C

Explanation

(3) Let `x_(1),x_(2),andx_(3)` be the abscissae of the points on the parabola whose ordinates are `y_(1),y_(2),andy_(3)`, respectively. Then `y_(1)^(2)=4ax_(1),y_(2)^(2)=4ax_(2),andy_(3)^(2)=4ax3`. Therefore, the area of the triangle whose vertices are `(x_(1),y_(1),(x_(2),y_(2)),and(x_(3),y_(3))` is
`Delta=(1)/(2){:||(x_(1)" "y_(1)" "1),(x_(2)" "y_(2)" "1),(x_(3)" "y_(3)" "1)||=(1)/(2)||(y_(1)^(2)//4a" "y_(1)" "1),(y_(2)^(2)//4a" "y_(2)" "1),(y_(3)^(2)//4a" "y_(3)" "1)||=(1)/(8a)||(y_(1)^(2)" "y_(1)" "1),(y_(2)^(2)" "y_(2)" "1),(y_(3)^(2)" "y_(3)" "1)||`
`=(1)/(8a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`