If there are two symmetric points on the ellipse x ^ 2 + 4Y ^ 2 = 64 about P (1,2), then the equation of line AB is (thank you)

If there are two symmetric points on the ellipse x ^ 2 + 4Y ^ 2 = 64 about P (1,2), then the equation of line AB is (thank you)

Let a (x1, Y1), B (X2, Y2)
Then X1 + x2 = 2, Y1 + y2 = 4
∵ a, B are on the ellipse,
∴ x1²+4y1²=64 ①
x2²+4y2²=64 ②
①-②
(x1²-x2²)+4(y1²-y2²)=0
∴ (x1-x2)(x1+x2)+4(y1-y2)(y1+y2)=0
2(x1-x2)+8(y1-y2)=0
∴ k(AB)=(y1-y2)/(x1-x2)=-2/8=-1/4
The linear AB: Y-2 = - (1 / 4) (x-1)
It is reduced to x + 4y-9 = 0

Given a point P (x, y) on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), find the value range of 3x = 4Y

It can be solved by the parameter equation of ellipse
The parameter equation of ellipse: x = a * COSC, y = b * sinc
3x + 4Y = a * COSC + b * sinc = radical (a ^ 2 + B ^ 2) sin (c + C2), where C2 is determined by tanc2 = A / b
So, the range is: [root (a ^ 2 + B ^ 2), root (a ^ 2 + B ^ 2)]

If the moving point P (x, y) satisfies a √ (x-1) 2 + (Y-2) 2 = | 3x + 4y-10 |, and the trajectory of P is ellipse, then the value range of a is larger
Under the root sign of a {the square of (x-1) plus the square of (Y-2)} = | 3x + 4y-10|

Eccentricity e = (distance to fixed point) / (distance to fixed line)
On the left √ (x-1) 2 + (Y-2) 2 is the distance to (1,2)
On the right | 3x + 4y-10 | add another 5 to get to the line 3x + 4y-10 = 0
So e = 5 / A
The e of ellipse is between (0,1)
So a > 5