Given the ellipse with foci at (5,3) and (5,-5) and a major axis length of 12, determine the Center and the denominators of the equation:                 Center - (h,k)              ...

Given the ellipse with foci at (5,3) and (5,-5) and a major axis length of 12, determine the Center and the denominators of the equation:

                Center - (h,k)

                (x  )^2  (y  )^2 = 1

Determine the denominators of the above equation

 

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Given two fixed points (foci) `F_1(5,3) and F_2(5,-5)` .Observe ordinates of fixed points are different ,it means major axis lies parallel to y -axis .

Equation of the major axis is

x=5

Center of the ellipse is mid point of `F_1 and F_2`  i.e.`((5+5)/2,(3-5)/2)=(5,-1)`

the length of major axis =12

Let (x,y) be the point on the ellipse . Thus By def. of the ellipse ,we have ,

`sqrt((x-5)^2+(y-3)^2)+sqrt((x-5)^2+(y+5)^2)=12`

`sqrt((x-5)^2+(y-3)^2)=12-sqrt((x-5)^2+(y+5)^2)`

squaring both side and simplify ,we get

`-2y-20=-3sqrt((x-5)^2+(y+5)^2)`

`` squaring again and simplify, we have

`5(-y^2-2y+35)=9(x-5)^2`

`-(y+1)^2+36=(9(x-5)^2)/5`

`(9(x-5)^2)/5+(y+1)^2=36`

`(x-5)^2/20+(y+1)^2/36=1`

 

 

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The ellipse has its foci at (5, 3) and (5,-5).

The foci have different y-coordinates indicating that the foci lie in the y-axis. Therefore, it is a vertical ellipse, having its major axis parallel to the y-axis. The standard form of the equation of the ellipse is:

`(x-h)^2/b^2+(y-k)^2/a^2=1`

with its center at (h, k), and  a ≥ b > 0.

For a vertical ellipse, coordinates of the two foci are (h, k+c) and (h, k-c), where c is the focal distance given by `c=sqrt(a^2-b^2)` . Hence the y-coordinate of the center will be the mid-point of the y-coordinates of the two foci.

Thus, `k=(3+(-5))/2=-1`

Center of the ellipse is at (5, -1).

Comparing the coordinates of the foci, we get the equations:

k+c=3 and k-c=-5, where k=-1

Upon solving, c=4.

Again, the length of the major axis, 2a=12

`rArr a=6`

Putting this value of a in the expression for c, we get

`4=sqrt(6^2-b^2)`

`rArr 16=36-b^2`

`rArr b^2=20`

Therefore, the required equation of the ellipse is:

`(x-5)^2/20+(y-(-1))^2/36=1`

`rArr (x-5)^2/20+(y+1)^2/36=1`

Sources:

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