Question

Write the equation of the ellipse with a minor axis length of 8 units and foci at (-7, 4) and (7, 4).

Answer 1

Step-by-step explanation:

x^2/65+(y-4)^2/16=1

Answer 2

Given:

Minor axis length is 8 units.

Focci of elipse : (-7,4) and (7,4).

Required:

To find the equation of the elipse.

Step-by-step explanation:

The general equation of elipse is

`\begin{gathered} (x^2)/(a^2)+(y^2)/(b^2)=1 \\ a>b \end{gathered}`

It is given that minor axis length is 8 units.

Therefore the value of b becomes 4.

Now we have to find the value of a.

And given that the focci points are (-7,4) and (7,4).

Here consider

`c=7`

Now,

`\begin{gathered} c^2=a^2-b^2 \\ 7^2=a^2-4^2 \\ 49=a^2-16 \\ 49+16=a^2 \\ a^2=65 \\ a=√(65) \end{gathered}`

Therefore, the equation of the elipse becomes

`(x^2)/(65)+((y-4)^2)/(16)=1`

Final Answer:

`(x^(2))/(65)+((y-4)^2)/(16)=1`