It is known that the center of the ellipse with eccentricity of 4 / 5 is at the origin and the focus is on the x-axis It is known that the center of the ellipse with eccentricity of 4 / 5 is at the origin and the focus is on the x-axis. The hyperbola takes the major axis of the ellipse as the real axis and the minor axis as the imaginary axis. If the focal length of the hyperbola is 2 root sign 34, the equations of the ellipse and hyperbola can be solved

It is known that the center of the ellipse with eccentricity of 4 / 5 is at the origin and the focus is on the x-axis It is known that the center of the ellipse with eccentricity of 4 / 5 is at the origin and the focus is on the x-axis. The hyperbola takes the major axis of the ellipse as the real axis and the minor axis as the imaginary axis. If the focal length of the hyperbola is 2 root sign 34, the equations of the ellipse and hyperbola can be solved

C / a = 4 / 5 let the hyperbolic equation x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
Elliptic equation x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
C (hyperbola) = root 34 A ^ 2 + B ^ 2 = C (hyperbola) ^ 2
The hyperbolic equation is x ^ 2 / 25-y ^ 2 / 9 = 1, and the elliptic equation is x ^ 2 / 25 + y ^ 2 / 9 = 1

It is known that the center of the ellipse C is at the origin o, the focus is on the x-axis, the length of its long axis is twice of the focal length, and it passes through the point (1,3 / 2). First question: find the standard equation of the ellipse C
Second question: if the line L with slope 1 intersects the ellipse at two different points a and B, find the maximum area of the triangle AOB and the equation of the line L at this time
I'm a sophomore in high school. I hope the steps are as complete as possible,

1. Let the elliptic equation be X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a > b > 0), C & # 178; = A & # 178; - B & # 178; (c > 0)
A = 2C ｜ B = radical 3 * C ｜ X & # / 4C & # / 178; + Y & # / 3C & # / 178; = 1
Substituting the point (1,3 / 2) into the equation, we get C & # 178; = 1. The elliptic standard equation is X & # 178 / 4 + Y & # 178 / 3 = 1
2. Let l be y = x + m, a (x1, Y1), B (X2, Y2), and l cross X-axis at C (- m, 0)
If y = x + m and X & # 178 / 4 + Y & # 178 / 3 = 1, then 7Y & # 178 / 12 - my / 2 + M & # 178 / 4 - 1 = 0,
According to Weida's theorem, Y 1 + y 2 = 6m / 7, Y 1 * y 2 = (3m & # 178; - 12) / 7
|Y1-y2 | = radical (48 / 7-48m & # / 49)
S△AOB=0.5 |y1-y2|*|-m|
S²△AOB=0.25（48/7-48m²/49）*m²=-12m^4 /49 +12m²/7=-12/49（m²-7/2）²+3
When M & # 178; = 7 / 2, S & # 178; △ aobmax = 3, s △ AOB = root 3
In this case, M = ± radical 14 / 2  y = x ± radical 14 / 2
If you are not clear, you are welcome to ask~

It is known that the center of the ellipse e is at the origin, the focus is on the x-axis, the focal length of the ellipse is 2, the eccentricity e = 1 / 2, and the straight line L: y = K (x-1) (K ≠ 0)
It is known that the center of the ellipse e is at the origin, the focus is on the x-axis, the focal length of the ellipse is 2, the eccentricity e = 1 / 2, and the straight line L: y = K (x-1) (K ≠ 0) intersects with the ellipse e at two points P and Q
(1) The equation for finding ellipse e
(2) Find the range of intercept of vertical bisector of line PQ on y-axis
Urgent need!

(1) The ellipse parameters can be determined by the following three formulas:
2C = 2 (focal length definition)
E = C / a = 1 / 2 (definition of eccentricity)
A ^ 2 = B ^ 2 + C ^ 2 (parameter relation)
The solution is a ^ 2 = 4, B ^ 2 = 3
So the ellipse e: x ^ 2 / 4 + y ^ 2 / 3 = 1
(2) Let P (x1, Y1), q (X2, Y2)
Substituting the linear l equation into the elliptic e equation is (3 + 4K ^ 2) x ^ 2-8k ^ 2x + 4K ^ 2-12
According to Weida's theorem, x 1 + x 2 = 8K ^ 2 / (3 + 4K ^ 2) (I)
Because P and Q are on the line L, then there is
y1=kx1-k
y2=kx2-k
By adding the two formulas and combining (I), we get Y1 + y2 = K (x1 + x2) - 2K = - 6K / (3 + 4K ^ 2) (II)
The PQ midpoint coordinates are obtained from the midpoint coordinates formula combined with (I) (II)
(x1+x2)/2=4k^2/(3+4k^2),(y1+y2)/2=-3k/(3+4k^2)
It is easy to know that the slope of PQ vertical bisector is - 1 / K
The equation of PQ vertical bisector is y = - (1 / k) x + M
Since the midpoint of PQ is on the vertical bisector of PQ, the coordinate satisfies the equation:
-3k/(3+4k^2)=-(1/k)*[4k^2/(3+4k^2)]+m
4 MK ^ 2-k + 3 M = 0
If M = 0, then k = 0
So m ≠ 0, and K exists, then ⊿ = 1-48m ^ 2 ≥ 0
The solution is - √ 3 / 12 ≤ M

Elliptic equation with center at origin, focus on y-axis, focal length of 8 and passing through point (3,0)
Is there a solution process? thank you

∵ the center is at the origin and the focus is on the y-axis
And passing through point (3,0)
∴b=3
∵ focal length is 8, ∵ C = 4
∴a^2=b^2+c^2=25
The elliptic equation: x ^ 2 / 9 + y ^ 2 / 25 = 1