Respuesta :
check the picture below.
[tex]\bf \begin{cases} h=0\\ k=0\\ a=9\\ b=4 \end{cases}\implies \cfrac{(x-0)^2}{9^2}+\cfrac{(y-0)^2}{4^2}=1\implies \cfrac{x^2}{81}+\cfrac{y^2}{16}=1[/tex]
[tex]\bf \begin{cases} h=0\\ k=0\\ a=9\\ b=4 \end{cases}\implies \cfrac{(x-0)^2}{9^2}+\cfrac{(y-0)^2}{4^2}=1\implies \cfrac{x^2}{81}+\cfrac{y^2}{16}=1[/tex]
Answer:
Equation of Ellipse in standard form is [tex]\frac{(x-0)^2}{81}+\frac{(y-0)^2}{16}=1[/tex]
Step-by-step explanation:
Given: Center of ellipse is ( 0 , 0 )
Ellipse is 8 units high.
⇒ Length of minor axis = 8
⇒ b = [tex]\frac{8}{2}=4[/tex]
Ellipse is 18 units wide.
⇒ Length of minor axis = 18
⇒ a = [tex]\frac{18}{2}=9[/tex]
Standard equation of ellipse whose major axis ia x-axis is given by,
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
where ( h , k ) is coordinates of center.
⇒ Equation of Ellipse : [tex]\frac{(x-0)^2}{9^2}+\frac{(y-0)^2}{4^2}=1[/tex]
[tex]\frac{x^2}{81}+\frac{y^2}{16}=1[/tex]
Therefore, Equation of Ellipse in standard form is [tex]\frac{(x-0)^2}{81}+\frac{(y-0)^2}{16}=1[/tex]