Respuesta :
Answer:
[tex]\frac{x^2}{169} +\frac{y^2}{25}=1[/tex]
If we compare this to the general expression for an ellipse given by:
[tex]\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2}=1[/tex]
We can see that the vertex is [tex] V=(0,0)[/tex]
And we can find the values of a and b like this:
[tex] a=\sqrt{169}=13, b=\sqrt{25}=5[/tex]
in order to find the foci we can find the value of c
[tex] c =\sqrt{169-25}=\sqrt{144}=12[/tex]
The two focis are (12,0) and (-12,0)
The convertices for this case are: (13,0) and (-13,0) on the x axis
And for the y axis (0,5) and (0,-5)
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 to the general expression for an ellipse given by:
[tex]\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2}=1[/tex]
We can see that the vertex is [tex] V=(0,0)[/tex]
And we can find the values of a and b like this:
[tex] a=\sqrt{169}=13, b=\sqrt{25}=5[/tex]
in order to find the foci we can find the value of c
[tex] c =\sqrt{169-25}=\sqrt{144}=12[/tex]
The two focis are (12,0) and (-12,0)
The convertices for this case are: (13,0) and (-13,0) on the x axis
And for the y axis (0,5) and (0,-5)
