Answer :
Answer:
The points that are the approximate locations of the foci of the ellipse are (0,√13) and (0,-√13) or, what is the same, (0,3.60) and (0,-3.60)
Step-by-step explanation:
The equation [tex]\frac{(y-8)^{2} }{49} +\frac{(x-1)^{2} }{36} =1[/tex] follows the following form of representation of an ellipse:[tex]\frac{(y-y_{0} )^{2} }{a^{2} } +\frac{(x-x_{0} )^{2} }{b^{2} } =1[/tex]
where x₀ and y₀ are the values of x and y in the center of the ellipce and a nab b are the values of the major and minor semiaxis, respectively. So the major axis is vertical in this case.
Being c the focal length, the relationship between the focal length and the semi-axes is: [tex]a^{2} =b^{2} +c^{2}[/tex] The foci are located on the major axis
In this case: 49=36+c²
Solving: c²=49-36
c²=13
c=√13 ≅ 3.60
So, the points that are the approximate locations of the foci of the ellipse are (0,√13) and (0,-√13) or, what is the same, (0,3.60) and (0,-3.60)