If the two vertices of the ellipse x2a2 + y2b2 = 1 (a > b > 0) are a (a, 0), B (0, b), and the left focus is f, △ fab is a right triangle with angle B as the right angle, then the eccentricity e of the ellipse is () A. 3-12B. 1+54C. 5-12D. 3+14

According to the meaning of the question, the slope of the straight line AB at f (- C, 0) is b-00-a = - BA, the slope of the straight line BF is 0-b-c-0 = BC ∵ FBA = 90 °, and the slope of the straight line BF is 0-b-c-0 = BC ∵ FBA = 90 °, and the slope of the straight line AB at f (- C, 0) is C2 + ac-a2 = 0, that is, (CA) 2 + CA-1 = 0, that is, e2-e-1 = 0, and the solution is e = 5-12 or 5 + 12 ∵ e < 1 ∵ e = 5-12

Given that the line y = KX + 2 and the ellipse x2 + y2 = 2 intersect at two different points, find the value range of K

Substituting y = KX + 2 into x2 + y2 = 2, we get: (K & sup2; + 1) x & sup2; + 4kx + 2 = 0
Δ=16k²-4(k²+1)2=8k²-8>0===>k²>1
Ψ k > 1 or

It is known that there must be a common point between the straight line y = kx-1 and the ellipse x ^ 2 / 3 + y ^ 2 / k = 1

According to the meaning of the topic, there is a common point between the line and the ellipse, which is only tangent,
y=kx-1
x^2/3 +y^2/k =1
Two style simultaneous
（1/3+k）x^2-2x+(1/k-1)=0
If the equation has a single root
The discriminant = 4-4 (1 / 3 + k) (1 / k-1) = 0 is 3K ^ 2 + K + 1 = 0
The equation has no real root, that is to say, no K value satisfies the problem meaning, so there is no solution

If the two vertices of the ellipse x2a2 + y2b2 = 1 (a > b > 0) are a (a, 0), B (0, b), and the left focus is f, △ fab is a right triangle with angle B as the right angle, then the eccentricity e of the ellipse is () A. 3-12B. 1+54C. 5-12D. 3+14