Question

NO LINKS!! Please help me with this problem
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Answer 1

`{\qquad\qquad\huge\underline{{\sf Answer}}}`

Here we go ~

The ellipse shown here is a horizontal ellipse that has major axis parallel to x - axis and minor axis parallel to y - axis. Length of major axis = 2a, and that of minor axis = 2b.

And it has it's centre on origin, it's equation can be written as :

`\qquad \sf  \dashrightarrow \: \cfrac{ {x}^(2) }{ {a}^(2) } + \cfrac{ {y}^(2) }{ {b}^(2) } = 1`

so, let's equate given equation with the standard equation ~

`\qquad \sf  \dashrightarrow \: \cfrac{ {x}^(2) }{64} + \cfrac{ {y}^(2) }{36} = 1`

so we get :

• a² = 64 ; a = 8

• b² = 36 ; b = 6

As we know, length of major axis is : 2a = 2 × 8 = 16 units

and, the larger circle has diameter = 2a = 16 units

so, it's radius = 8 units ~

Now, let's write the equation of circle with origin as centre and radius = 8 units

`\qquad \sf  \dashrightarrow \: {(x - h)}^(2) + (y - k) {}^(2) = {r}^(2)`

[ h = 0, k = 0, since circle has centre at origin ]

`\qquad \sf  \dashrightarrow \: {x}^(2) + {y}^(2) = 64`

Answer 2

`x^2+y^2=64`

Explanation:

So the first important thing in solving this problem, is identifying what the major and minor axis are. The major axis is the bigger one, and we want to find on which axis it is.

So by simply looking at the eclipse, you can see that it's larger on the horizontal axis, so the major axis is on the horizontal axis, and the minor axis is on the vertical axis.

This means the equation will be expressed as:

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

In this equation the major axis, has a length of 2a, and the minor axis has a length of 2b. It's also important to note that (h, k) is the middle of the eclipse, but in the equation you provided, there is no subtraction, so it's just 0, meaning the center is at the origin (0, 0)

So now let's solve for a, and b. I'll look at each individual fraction separately.

`(x^2)/(64)`

The denominator is equal to the value of a^2, so we simply take the square root of this to get the equation: a=8

`(y^2)/(36)`

The denominator is equal to the value of b^2, so we simply take the square root of this to get the equation: b=6

So by just looking at the graph the larger circle appears to have a denominator that is equal to the length of the major axis. The major axis is equal to 2a, and since we know the value of a (8), we simply multiply this by 2 to get a length of 16. This was a bit redundant to do, since in the equation of a circle we need the radius, and the radius is just half the diameter, so now we divide this 16 by 2 to get a radius of 8.

The circle also appears to be centered at the origin so the equation will have (x-0)^2 and (y-0)^2 which is just x^2 and y^2

Plugging in all the values we get the equation:

`x^2+y^2=64`