Saved Bookmarks
| 1. |
y^(2)=4xandy^(2)=-8(x-a) intersect at points A and C. Points O(0,0), A,B (a,0), and c are concyclic. Tangents to the parabola y^(2)=4x at A and C intersect at point D and tangents to the parabola y^(2)=-8(x-a) intersect at point E. Then the area of quadrilateral DAEC is |
|
Answer» `96sqrt(2)` Solving the fiven parabolas , we have -8(x-1)=4x `orx=(2A)/(3)` Therefore, the INTERSECTION are `(2a//3,pmsqrt(8a//3))`. Now, OABC is cyclic quadrilateral. Hence, `angleOAB` must be a right angle. So, Slope of `OAxx` Slope of AB=-1 `or(sqrt(8a//3))/(2a//3)xx(sqrt(8a//3))/(a-(2a//3))=-1` `ora=12` Therefore, the coordinates of A and B are `(8,4sqrt(2))and(8,-4sqrt(2))`, respectively. So, Length of COMMON chord `=8sqrt(2)` Area of quadrilateral `=(1)/(2)OBxxAC` `=(1)/(2)xx12xx8sqrt(2)` `48sqrt(2)` Tangent to the parabola `y^(2)=4xat(8,4sqrt(2))" is "4sqrt(2)y=2(x+8)orx-2sqrt(2)y+8=0`, which meets the x-axis at D(-8,0). Tangent to the parabola `y^(2)=-8(x-12)at(8,4sqrt(2))" is "4sqrt(2)y=-4(x+8)+96orx+sqrt(2)y=16=0`, which meets the x-axis at E(16,0). Hence, Area of quadrilateral `DAEC=(1)/(2)DExxAC` `(1)/(2)xx24xx8sqrt(2)` `=96sqrt(2)` |
|