Answer:
Step-by-step explanation:
If you plot those points on a coordinate plane, you'll see that the distance from the origin up the y-axis to the point is greater than is the distance from the origin down the x-axis to the other point. That means 3 things to us: 1. the greater distance is a and the shorter is b; 2. the point (0, 11) is the vertex while the point (4, 0) is the co-vertex; and 3. this is a vertically stretched ellipse. A vertically stretched ellipse has an equation
[tex]\frac{(x-h)^2}{b^2} +\frac{(y-k)^2}{a^2} =1[/tex] where h and k are the coordinates of the center, a is the greater distance (between the center and the vertex), and b is the smaller distance (between the center and the co-vertex). Here's what we have then thus far:
h = 0
k = 0
a = 11
b = 4
Filling in our equation then looks like this:
[tex]\frac{(x-0)^2}{4^2} +\frac{(y-0)^2}{11^2} =1[/tex] and simplifying:
[tex]\frac{x^2}{16} +\frac{y^2}{121} =1[/tex]. It appears that the last answer is the one you want, although when I teach this to my precalc students, I do not encourage them to move the x and y terms around as that answer appears to have done. But addition is also commutative so I'm sure it's acceptable (I just think it looks strange that way).
The equation of ellipse is [tex]\dfrac{y^2}{121}+\dfrac{x^2}{16}=1[/tex] option D is correct.
Important information:
The standard form of an ellipse is:
[tex]\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1[/tex]
Where, [tex](0,a)[/tex] is vertex and [tex](0,b)[/tex] is vertex.
Substitute [tex]a=11,b=4[/tex] in the above equation.
[tex]\dfrac{x^2}{(4)^2}+\dfrac{y^2}{(11)^2}=1[/tex]
[tex]\dfrac{x^2}{16}+\dfrac{y^2}{121}=1[/tex]
[tex]\dfrac{y^2}{121}+\dfrac{x^2}{16}=1[/tex]
Therefore, the correct option is D.
Find out more about 'Ellipse' here:
https://brainly.com/question/1548816
