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If `y_(1),y_(2),andy_(3)` are the ordinates of the vertices of a triangle inscribed in the parabola `y^(2)=4ax`, then its area isA. `(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 |
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Answer» Correct Answer - C (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))|` |
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