Given the point P (x, y) on the ellipse {x = ACOS Φ;} {y = bsin Φ} (a ＞ B ＞ 0, Φ is a parameter), find: 1) the value range of X, y; 2) the value range of 3x + 4Y

Given the point P (x, y) on the ellipse {x = ACOS Φ;} {y = bsin Φ} (a ＞ B ＞ 0, Φ is a parameter), find: 1) the value range of X, y; 2) the value range of 3x + 4Y

Cos Φ, sin Φ in [- 1,1], so - A

Given the point P (x, y) on the ellipse x2 / 16 + Y2 / 9 = 1, find the value range of 3x + 4Y

Let 3x + 4Y = K,
x2/16+y2/9=1
If a line and an ellipse have intersection points, join them, and the discriminant is greater than or equal to 0, the range of K is obtained

2X square + y square = 8 3x square + 4Y square = 12

As a result, we will be able to get the final result of the 178; + y-178; + y-178; + y-178; = 8x-178; (4 + y-178; = 8, b-178; = 4, c-178; = 8-4 = 4, c-178; = 8-4 = 4, c-178; = 8-4 = 4, c-178; = 8-4 = 4, C-4 = 4, C = 2 focus: (0, - 2), (0,2), focal length = 2C = 43x-43x-178; (4-43x-178; + 4xy-178; (4x-178; = 12x-178; (12x-178;; 12x-178; (4, and this will be the 178; (4, b-178;; 12x-178; (4; 4, 4, b-178; 4, b-y-178; = 12x-178; 4, we will be the 178; 4, b-178; (4 + y-178; 4, b-178; (4, b-178 1

Find the elliptic standard equation with common focus between the passing point (√ 2, - 2 √ 15 / 2) and the ellipse 3x ^ 2 + 4Y ^ 2 = 12

Ellipse 3x ^ 2 + 4Y ^ 2 = 12
Ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1
c^2=4-3=1
Let the elliptic equation be:
x^2/a^2+y^2/b^2=1
Because there is a common focus, so
a^2-b^2=1
It's a little over (√ 2, - 2 √ 15 / 2), so
2 / A ^ 2 + 30 / b ^ 2 = 1
The solution
a^2=

Find the focus and focal length of ellipse x Λ + 4Y Λ 2 = 4

X ^ 2 + 4Y ^ 2 = 4 can be obtained by dividing by 4
x^2/4+y^2=1
So, the focus is on the x-axis
So, a ^ 2 = 4, B ^ 2 = 1
It can be obtained from the formula A ^ 2 = B ^ 2 + C ^ 2
c^2=4-1=3
c=√3
So the focus is (√ 3,0) and (- √ 3,0)
Focal length = 2C = 2 √ 3