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The focal points F1 (- 5,0) and F2 (5,0) of hyperbola C are known, and the length of major axis is 6 (1) Let the line y = x + 2 intersect the hyperbola C at two points a and B, and find the midpoint coordinates of line ab. (2) if the line y = KX + 2 has an intersection with the hyperbola, find the value range of K. (3) if the line y = KX + 2 has two intersections with a branch of the hyperbola, and the intersection points are a and B respectively, find the value range of K
(1) C = 5, real axis length 2A = 6, a = 3,
∴b^2=c^2-a^2=16,
The hyperbola C: x ^ 2 / 9-y ^ 2 / 16 = 1
Let y = x + 2, 2
Substituting ① * 144, 16x ^ 2-9 (x ^ 2 + 4x + 4) = 144,
The results show that 7x ^ 2-36x-180 = 0,
The coordinates of the midpoint m of line AB: x = (x1 + x2) / 2 = (36 / 7) / 2 = 18 / 7, substituting into 2, y = 32 / 7, i.e. m (18 / 7,32 / 7)
(2) Substituting y = KX + 2 into ① * 144,16x ^ 2-9 (k ^ 2x ^ 2 + 4kx + 4) = 144,
(16-9k^2)x^2-36kx-180=0,③
According to the meaning of 16-9k ^ 2 = 0, or △ / 16 = 81k ^ 2 + 45 (16-9k ^ 2) = 0,
The solution is k = soil 4 / 3, or soil 2 √ 5 / 3
(3) K ^ 20, K ^ 2 can be obtained from △ / 16 > 0
A focus coordinate of hyperbola is (- 5,0), and the real axis length is 8
According to the meaning of the title
c=5,2a=8,a=4
The focus is on the x-axis
And B & sup2; = C & sup2; - A & sup2; = 5 & sup2; - 4 & sup2; = 9
So the equation: X & sup2 / 16-y & sup2 / 9 = 1
It is known that a straight line y = X-1 and an ellipse x2m + y2m − 1 = 1 (M & gt; 1) intersect at two points a and B. If a circle with diameter AB passes through the left focus F of the ellipse, then the value of real number m is___ ．
From the topic meaning, C = A2-B2 = 1, f (- 1, 0) line y = X-1, the ellipse x2m + y2m-1 = 1 is substituted into the ellipse x2m + y2m-1 = 1, and then (2m-1) x2-2mx + 2m-m2 = 0, let a (x1, Y1, Y1) and B (X2, Y2), then X1 + x2 = 2m2m-1, X (- 1, 0) line y = X-1, the ellipse x2m + y2m-2m-1 = 1 = 1, and (2m-1) (x2-1) (x2-1) (x2-1) (x2-1) = (x1-1) (x2-1) (x2-1) = (x1-1) (x2-1) (x2-1) = (x2-1) (x2-1) (x2-1) = (x2-m2 + 2m-2 + 2m-2m-2m-2m-2m-2m-12m-2m-2m-2m-2m-2m-12m-2m-12m-12-1- 1 +1 + - M2 + 2m-12m-1 = 0 ∫ m2-4m + 1 = 0 ∫ M = 2 ± 3 ∵ M & gt; 1 ∫ M = 2 + 3, so the answer is: 2 + 3
Given that the line y = x + 1 and the ellipse x ^ 2 / M + y ^ 2 / M-1 (M > 1) intersect at points a and B, if the circle with diameter AB just passes through the left focus F of the ellipse, find the value of the real number M
Because a ^ 2 = m, B ^ 2 = M-1, so C = 1, so the line y = x + 1 itself is a straight line passing through the left focus F1 (- 1,0) and the point (0,1). It and the two focuses of the ellipse are on both sides of F1. How can a circle with the diameter of these two points pass through F1? If it just passes through the right focus F2 (1,0), it can be solved by using the second definition, Pythagorean theorem and Weida theorem,
|AF2|^2+|BF2|^2=(|AF1|+|F1|)^2
That is, (a-ex1) ^ 2 + (a-ex2) ^ 2 = [(a + EX1) + (a + ex2)] ^ 2
It is reduced to m ^ 2 + 3M (x1 + x2) + x1x2 = 0
Substituting y = KX + 1 into the elliptic equation, we can get the result by using the Veda theorem
x1+x2=2m/(1-2m),x1x2=m(m-2)/(1-2m) ②
② We can get m = 2 + 3 ^ (1 / 2)
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