The ellipse with focus on X axis X ^ 2 / 4A + y ^ 2 / (a ^ 2 + 1) = 1, and the maximum eccentricity is

The ellipse with focus on X axis X ^ 2 / 4A + y ^ 2 / (a ^ 2 + 1) = 1, and the maximum eccentricity is

The focus is on the x-axis
Then E & # 178; = (4a-a & # 178; - 1) / 4A = 1-1 / 4 / (a + 1 / a)
a+1/a≥2√(a*1/a)=2
Take the equal sign when x = 1
In this case 4A > A & # 178; + 1
In line with the theme
So E & # 178; ≤ 1-1 / 4 * 2 = 1 / 2
So the maximum value of E is √ 2 / 2

If the focus of the ellipse x ^ 2 / 5A + y ^ 2 / 4A + 1 = 1 is on the x-axis, then the ellipse is eccentric

The focus is on the X axis
5a>4a+1
a>1
c²=5a-4a-1=a-1
e²=c²/a²=(a-1)/5a
5e²=(a-1)/a=1-1/a
a>1
0

If the equation x2a2 − y2a = 1 represents an ellipse with focus on the y-axis, then the value range of a is ()
A. - 1 < a < 0b. 0 < a < 1C. A < 1D

According to the meaning of the problem, the equation x2a2 − y2a = 1, that is, x2a2 + Y2 − a = 1, which means that the ellipse with focus on the y-axis ∧ - a ∧ A2 ∧ 0, ∧ - 1 ∧ a ∧ 0, so a

If P is any point on the ellipse x2a2 + y2b2 = 1 (a > b > 0), P is perpendicular to the two focal lines, and the distances from P to the two collimators are 6 and 12 respectively, then the ellipse equation is ⊙___ ．

Since the distance from P to the two collimators is 6 and 12 respectively, let p to the left collimator be 6, then 12 + 6 = 2a2c, that is, A2C = 9. Because the ratio of the distance from the point on the ellipse to the focus and the distance to the collimator is eccentricity e, so | Pf1 | is 6e, | PF2 | is 12e. Because Pf1 is perpendicular to PF2, so | F1F2 | 2 = (6e) 2 + (12e) 2 = 180e2 = 4C2, so A2 = 45, C = 5 is obtained from A2C = 9, | B2 = a2-c2 = 20, so the ellipse is square The course is x245 + y220 = 1, so the answer is x245 + y220 = 1