Respuesta :
Answer:
We can see that for this case the vertex is [tex] V= (0,0)[/tex]
The values for a and b are:
[tex] a = \sqrt{169}=13, b= \sqrt{25}=1[/tex]
Then the ellipse have the major axis on x.
In order to find the two foci we need to use the following formula:
[tex] c =\sqrt{a^2 -b^2}[/tex]
And replacing we got:
[tex] c =\sqrt{169-25}= \pm 12[/tex]
And then the two foci are (12,0) and (-12,0)
And the covertex are on this case (-13,0) (13,0) and (0,5) (0,-5) on the y axis
Step-by-step explanation:
For this problem we have the following equation given:
[tex]\frac{x^2}{169} + \frac{y^2}{25}= 1[/tex]
If we compare this with the general expression of an ellipse given by:
[tex] \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}= 1[/tex]
We can see that for this case the vertex is [tex] V= (0,0)[/tex]
The values for a and b are:
[tex] a = \sqrt{169}=13, b= \sqrt{25}=1[/tex]
Then the ellipse have the major axis on x.
In order to find the two foci we need to use the following formula:
[tex] c =\sqrt{a^2 -b^2}[/tex]
And replacing we got:
[tex] c =\sqrt{169-25}= \pm 12[/tex]
And then the two foci are (12,0) and (-12,0)
And the covertex are on this case (-13,0) (13,0) and (0,5) (0,-5) on the y axis
