It is known that the equation of hyperbola is x2a2-y2b2 = 1, points a and B are on the right branch of hyperbola, line AB passes through the right focus F2 of hyperbola, | ab | = m, F1 is another focus, then the perimeter of △ Abf1 is () A. 2a+2mB. a+mC. 4a+2mD. 2a+4m

It is known that the equation of hyperbola is x2a2-y2b2 = 1, points a and B are on the right branch of hyperbola, line AB passes through the right focus F2 of hyperbola, | ab | = m, F1 is another focus, then the perimeter of △ Abf1 is () A. 2a+2mB. a+mC. 4a+2mD. 2a+4m

∫ AF1 | - | af2 | = 2A, | BF1 | - | BF2 | = 2A, and | af2 | + | BF2 | = | ab | = m, | AF1 | + | BF1 | = 4A + m, the perimeter of | △ Abf1 = | AF1 | + | BF1 | + | ab | = 4A + 2 | ab | = 4A + 2m

What is the elliptic standard equation with the focus of F1 (0, - 6) and F2 (0,6) through the point a (2, - 5)

Question 1: let the elliptic equation be X & sup2 / B & sup2; + Y & sup2 / A & sup2; = 1
Substitute the coordinates of a into 4 / B & sup2; + 25 / A & sup2; = 1 ------ (1)
From the focal coordinate C = 6 ∥ A & sup2; - B & sup2; = 36 ------ (2)
Simultaneous (1) and (2) give a & sup2; = 20, B & sup2; = 16
So the elliptic equation is X & sup2 / 20 + Y & sup2 / 16 = 1
Question 2: let the hyperbolic equation be - X & sup2 / A & sup2; + Y & sup2 / B & sup2; = 1
Substituting the coordinates of a into - 4 / A & sup2; + 25 / B & sup2; = 1 - - (1)
From the focal coordinates, we know C = 6 ∥ A & sup2; + B & sup2; = 36 ------ (2)
Simultaneous (1) and (2) give a & sup2; = 16, B & sup2; = 20
So the elliptic equation is - X & sup2 / 16 + Y & sup2 / 20 = 1

On the standard equation of hyperbola
Find the standard equation of hyperbola satisfying the following conditions respectively
(1) The center of the hyperbola is at the origin, and the focus is on the coordinate axis, passing through two points (7,6 root sign 3), (2 root sign 7, - 3);
(2) It is known that hyperbola and x ^ 2-2y ^ 2 = 2 have a common asymptote, and one focus is f (3,0)

1\ 11x^2/287-y^2/123=1
2\ x^2/6-y^2/3=1

Solution of absolute value quadratic equation with one variable

For example, when ∣ x + 4 ∣ is treated, the range of X is divided into x less than - 4, X equal to - 4 and x greater than - 4, thus eliminating the absolute value, transforming the original equation into a general equation, and then solving it. Another example is when ∣ x ∣ 178; - 4 = 3 ∣ x ∣ 178; - 4 = 3 ∣ x ∣, If you are not familiar with the method of dividing the range, please refer to the encyclopedia "zero segmentation method"
For example, the solution of X ∣ 178; - 4 = 6 ∣ x ∣ x ∣; - 4 = 6 ∣ x ∣; - 4 = 6 ∣ x ∣ x 8739; 8739; 178; for example, the solution of X ∣ 178;, for example, the solution of X ∣ 178; as ∣ x ∣ 178; as ∣ x ∣ x ∣ as an unknown number to solve, the solution of ∣ x ∣ x ∣ x ∣ x ∣ x ∣ x ∣ there are three advantages, Image of the said, is the "bridge."
Of course, it is not necessary to use these two methods for some bivariate quadratic equations with absolute values (but these two methods are generally applicable). For some test questions with ingenious solutions (such as finding the number of solutions of bivariate quadratic equations with absolute values), we can observe and analyze the essence of the problem, and sometimes it is also a way of thinking, Therefore, the solution is not limited to the above two methods