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If the left focus of hyperbola x23 − 16y2p2 = 1 is on the Quasilinear of parabola y2 = 2px, then the value of P is () A. 2B. 3C. 4D. 42
The left focus coordinate of hyperbola is: (− 3 + P216, 0), the Quasilinear equation of parabola y2 = 2px is x = − P2, so − 3 + P216 = − P2, the solution is: P = 4, so choose C
What is the weight of the Quasilinear of the parabola y = 2px to the left quasilinear of the hyperbola X-Y = 2 and the focal coordinate of the parabola
A:
Parabola y ^ 2 = 2px
Hyperbola x ^ 2-y ^ 2 = 2, (x ^ 2) / 2 - (y ^ 2) / 2 = 1
So: A ^ 2 = B ^ 2 = 2
So: C ^ 2 = a ^ 2 + B ^ 2 = 4
The solution is: C = 2
So: the focus of hyperbola is (- 2,0), (2,0), and the Quasilinear equation is x = ± (a ^ 2) / C = ± 1
So: the Quasilinear equation of parabola is x = - P / 2 = - 1
So: P = 2
So: parabola is y ^ 2 = 2px = 4x, focus is (1,0)
The coordinates of the intersection of the line y = 2x and the hyperbola y = 1 / X are____
The two equations are simultaneous, 2x = 1 / x, x = root two of plus and minus two
Substituting y = 2x, y = positive and negative root sign two
Intersection coordinates
(x, y) are (2 / 2 radical 2, radical 2) and (2 / 2 negative radical 2, negative radical 2)
As shown in the figure, if one of the intersection coordinates of the image of the straight line y = 2x and the hyperbola y = KX is (2,4), then the other of their intersection coordinates is ()
A. （-2，-4）B. （-2，4）C. （-4，-2）D. （2，-4）
Since the inverse scale function is a centrosymmetric figure, the intersection points a and B of the positive scale function y = 2x and the inverse scale function y = KX are symmetric about the origin. Also, because the coordinates of the point (2,4) about the origin are (- 2, - 4), a
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