In Cartesian coordinates, the set of points in the second quadrant is

In Cartesian coordinates, the set of points in the second quadrant is

｛（x,y）｜（x＜0,y＞0）｝

Given that the center of the ellipse is at the origin, taking the coordinate axis as the symmetry axis, and passing through two points P & # 8321; (√ 6,1) P & # 8322; (- √ 3, # 2), the equation of the ellipse is solved

Let the elliptic equation be MX ^ 2 + NY ^ 2 = 1, through two points P &; (√ 6,1) P &; (- √ 3, - √ 2), and substitute it into 6m + n = 13m + 2n = 16m + 4N = 23n = 1, n = 1 / 3M = 1 / 9. The elliptic equation x ^ 2 / 9 + y ^ 2 / 3 = 1 can also be set as x ^ 2 / A ^ 2 + y ^ 2 / A ^ 2 = 1 (when the focus is on the X axis)

When the brackets are removed, are the symbols in front of the brackets removed?

Yes, if there is a negative sign in front of the bracket, then all the change sign "+" in the bracket will become "+", "-" will become "+"
If there is a positive sign in front of the bracket, the sign inside the bracket will not change after removing it

(1 / 2) ^ (log2 ^ 3)

(1/2)^(log2^ 3)
＝1／2＾（log2^3)
=1/3

As shown in the figure, the right angle side BC of RT △ ABC is on the positive half axis of X axis, the middle line BD on the hypotenuse side AC is opposite to the extension line of Y axis and the negative half axis of E, and the image of hyperbola y = KX (x > 0) passes through point A. if the area of △ BEC is 4, then K is equal to ()
A. 16B. 8C. 4D. 2

∵ BD is the middle line on the hypotenuse AC of RT △ ABC, ∵ BD = DC, ∵ DBC = ∵ ACB, ∵ DBC = ∵ EBO, ∵ EBO = ∵ ACB, ∵ BOE = ∵ CBA = 90 °, ∵ BOE ∵ CBA, ∵ Bobc = oeab, namely BC × OE = Bo × ab. s △ BEC = 4,