Respuesta :
Answer:
[tex]\frac{(x-1)^2}{81}+\frac{(y-6)^2}{72}=1[/tex]
Step-by-step explanation:
We have been given that the center of an ellipse is (1,6). One focus of the ellipse is (-2,6). One vertex of the ellipse is (10,6).
We know that standard equation of an ellipse centered at (h,k) is in form:
[tex]\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1[/tex], where,
a = Horizontal radius
b = Vertical radius.
Since center of the ellipse is at point (1,6) and one vertex is (10,6), so the horizontal radius would be 9 (10-1) units.
Now, we will find vertical radius using formula [tex]b^2=a^2-c^2[/tex], where c represents focal length.
[tex]\text{Focal length}=1-(-2)[/tex]
[tex]\text{Focal length}=1+2[/tex]
[tex]\text{Focal length}=3[/tex]
Substituting given values:
[tex]b^2=9^2-3^2[/tex]
[tex]b^2=81-9[/tex]
[tex]b^2=72[/tex]
Upon substituting [tex]a=9[/tex], [tex]b^2=72[/tex] and [tex](h,k)=(1,6)[/tex] in standard form of ellipse, we will get:
[tex]\frac{(x-1)^2}{9^2}+\frac{(y-6)^2}{72}=1[/tex]
[tex]\frac{(x-1)^2}{81}+\frac{(y-6)^2}{72}=1[/tex]
Therefore, the required equation of the ellipse in standard form would be [tex]\frac{(x-1)^2}{81}+\frac{(y-6)^2}{72}=1[/tex].
