Question

#45 please, just checking work. p.s. if i am slow to reply it is because i have been doing homework for 9 hours

Answer 1

Given:

Vertices:

`(5,0),(-5,0)`

Co-vertices:

`(0,4)(0,-4)`

To find: The ellipse equation

Step-by-step explanation:

The general form of ellipse equation is,

`((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1`

Where (h, k) is the center and a and b are the length of the semi-major and semi-minor axis.

Here,

`(h,k)=(0,0)`

The length of the major axis is the distance between the vertices.

`\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt[]{(0-0)^2+(-5_{}-5)^2} \\ =\sqrt[]{10^2} \\ =10\text{ units} \end{gathered}`

The length of the minor axis is the distance between the co-vertices.

`\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt[]{(-4-4)^2+(0-0)^2} \\ =\sqrt[]{8^2} \\ =8\text{ units} \end{gathered}`

Therefore,

The length of the semi-major axis is 5

The length of the semi-minor axis is 4

So, the standard form of an ellipse equation becomes,

`\begin{gathered} ((x-0)^2)/(5^2)+((y-0)^2)/(4^2)=1 \\ (x^2)/(5^2)+(y^2)/(4^2)=1 \end{gathered}`

Final answer:

The standard form of an ellipse equation becomes,

`(x^2)/(5^2)+(y^2)/(4^2)=1`