The product of the lengths of the real and imaginary semiaxes of the hyperbola is root sign 3, F1 and F2 are the left and right focal points, the straight line L passes through point F2, and the included angle with the straight line F1F2 is θ, Tan θ = root sign 21 / 2, the intersection of the vertical bisector of the straight line L and F1F2 with point P, the intersection of the line PF2 and the hyperbola with Q, and | PQ |: | QF2 | = 2:1, the hyperbolic equation is solved

The product of the lengths of the real and imaginary semiaxes of the hyperbola is root sign 3, F1 and F2 are the left and right focal points, the straight line L passes through point F2, and the included angle with the straight line F1F2 is θ, Tan θ = root sign 21 / 2, the intersection of the vertical bisector of the straight line L and F1F2 with point P, the intersection of the line PF2 and the hyperbola with Q, and | PQ |: | QF2 | = 2:1, the hyperbolic equation is solved

Suppose x ^ / A ^ - y ^ / b ^ = 1, q (x, y), F2 (C, 0), make X-axis vertical line through Q, vertical foot is a, PQ: QF2 = 2:1 = OA: AF, OA + AF = C, so: OA = 2C / 3 = x, af2 = C / 3, Tan α = (√ 21) / 2 = Y / AF = = = = = > y = (√ 21) C / 6, that is: Q (2C / 3, √ 21 C / 6) is substituted into the equation, 4C ^ / 9A ^ - 7C ^ / 12b ^ = 1, C ^ = a ^ + B ^

As shown in the figure, object a is in equilibrium under the action of common point forces F1, F2 and F3. Now turn F2 anticlockwise by 60 degrees (other forces are unchanged), and the resultant force on object a becomes ()
A. F3B. F1+F3C. 2F2D. F2

In force balance, the resultant force of any two forces and the third force are equivalent, reverse and collinear, so the resultant force of the two forces F1 and F3 except F2 is equal to F2, and the direction is opposite to F2, so it is equivalent to an object under the action of two forces with the size of 120 ° equal to F2. According to the rule of parallelogram, when two forces with the size of 120 ° are combined, the resultant force is on the angular bisector of the two components In this case, the resultant force on the object is F2, so D

If F1 = 10N, F2 = 20n, F3 = 15N, then the resultant force of F1 and F2 is equal

The resultant force of the object is 0 when three forces are in balance
So the resultant force of F1 and F2 is the opposite force of F3, with opposite direction and equal size
The resultant force of F1 and F2 is 15N

If F4 is 10N, then the resultant force of the other three forces is 10N（
）N. If the F4 direction turns 90 ° and its magnitude remains the same as 10N, and the magnitude and direction of the other three forces remain the same, then the resultant force on the object is () n

If we regard the combination of three forces as one force and the fourth force, they should be equal in magnitude and opposite in direction, so the first one asks 10N, and the second one turns around and the two forces become vertical, so it is 10 times the root sign 2