An ellipse and a hyperbola have their principal axes along the coordinate axes and have a common foci separated by distance `2sqrt(3)dot` The difference of their focal semi-axes is equal to 4. If the ratio of their eccentricities is 3/7 , find the equation of these curves.

Explanation

Let the semi-major axis, semi-minor axis, and eccentricity of the ellipse be a,b, and e, respectively, the those of hyperbola be a', b', and c', respectively.
Given that `a-a'=4" (1)"`
`2ae=2sqrt(13)or ae=sqrt(13)" (2)"`
`2a'e'=2sqrt(13)or a'e'=sqrt(13)" (3)"`
and `(e)/(e')=(3)/(7)" (4)"`
Form (2), `a=(sqrt(13))/(e)" (5)"`
Form (3), `a'=(sqrt(13))/(e')" (6)"`
Putting in (1), we get
`(sqrt(13))/(e)-(sqrt(13))/(e')=4`
`"or "sqrt(13)(e'-e)=4ee'`
Putting `e=3e'//7,` we get
`sqrt(13)(e'-(3)/(7)e')=4xx(3)/(7)(e')^(2)`
`therefore" "e'=(sqrt(3))/(3) and e=(3)/(7)e'=(sqrt(13))/(7)`
`a=(sqrt(13))/(e)=(sqrt(13))/(sqrt(13))xx7=7`
`a'=(sqrt(13))/(e')=(sqrt(13))/(sqrt(13))xx3=3`
`b^(2)=a^(2)(1-e^(2))=49(1-(13)/(49))=36`
`(b')^(2)=(a')^(2){(e')^(2)-1}=9((13)/(9)-1)=4`
The equation of ellipse is
`(x^(2))/(49)+(y^(2))/(36)=1`
and the equation of hyperbola is
`(x^(2))/(9)-(y^(2))/(4)=1`