[tex]\frac{x^2}{4} +\frac{y^2}{49} =1[/tex]
General equation is
[tex]\frac{(x-h)^2}{b^2} +\frac{(y-k)^2}{a^2} =1[/tex]
Where (h,k) is the center
From the given equation h=0 and k=0
So center is (0,0)
compare the given equation with general equation
b^2 = 4 so b= 2
a^2 = 49 so a = 7
[tex]c=\sqrt{a^2 -b^2}[/tex]
[tex]c=\sqrt{49 -4}=3\sqrt{5}[/tex]
Vertices are (h, k+a) and (h, k-a)
We know h=0 , k=0 and a= 7
Vertices are (0,-7) and (0,7)
Foci are (h, k+c) and (h,k-c)
We know h=0 , k=0 and c=[tex]3\sqrt{5}[/tex]
Foci are (0,-[tex]3\sqrt{5}[/tex]) and (0,[tex]3\sqrt{5}[/tex])
Answer with Step-by-step explanation:
We have to graph the ellipse:
[tex]\dfrac{x^2}{4}+\dfrac{y^2}{49}=1[/tex]
when x=0 , [tex]\dfrac{y^2}{49}=1[/tex]
i.e. y²=49
i.e. y= ±7
ellipse passes through (0,7) and (0,-7)
when y=0, [tex]\dfrac{x^2}{4}=1[/tex]
i.e. x² = 4
i.e. x= ± 2
i.e. ellipse passes through (2,0) and (-2,0)
The ellipse is shown as below:
