Question

NO LINKS!! Please help me with this problem.​

Answer 1

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

Information : The given ellipse is a horizontal ellipse, and it's centre lies on origin, as the foci are given on x - axis.

The foci of the ellipse can be written in form :

`\qquad \sf  \dashrightarrow \: ( \pm ae , 0)`

So,

`\qquad \sf  \dashrightarrow \: ae = 5`

and Vertex of the ellipse can be written as

`\qquad \sf  \dashrightarrow \: (\pm a,0)`

• so, we get a = 11

Now, plug the value in first equation ~

`\qquad \sf  \dashrightarrow \: ae = 5`

`\qquad \sf  \dashrightarrow \: 11e = 5`

`\qquad \sf  \dashrightarrow \: e = \cfrac{5}{11}`

Now, we have to find b (length of semi minor axis)

we can use the formula ~

`\qquad \sf  \dashrightarrow \: b {}^(2) = {a}^(2) (1 - {e}^(2) )`

`\qquad \sf  \dashrightarrow \: b {}^(2) = 121(1 - (25)/(121) )`

`\qquad \sf  \dashrightarrow \: b {}^(2) = 121( (121 - 25)/(121) )`

`\qquad \sf  \dashrightarrow \: b {}^(2) = 96`

Now, we can write the equation of ellipse as :

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

[ plug the values ]

`\qquad \sf  \dashrightarrow \: \cfrac{ {x}^(2) }{ {121}^{} } + \cfrac{ {y}^(2) }{ {96}^{} } = 1`

Answer 2

`(x^2)/(121)+(y^2)/(96)=1`

Explanation:

General equation of an ellipse:

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

where:

• center = (h, k)
• Vertices = (h±a, k) and (h, k±b)
• Foci = (h±c, k) and (k, h±c) where c²=a²−b²
• Major Axis: longest diameter of an ellipse
• Minor Axis: shortest diameter of an ellipse
• Major radius: one half of the major axis
• Minor radius: one half of the minor axis

If a > b the ellipse is horizontal, a is the major radius, and b is the minor radius.

If b > a the ellipse is vertical, b is the major radius, and a is the minor radius.

Given:

• foci = (-5, 0) and (5, 0)
• vertices = (-11, 0) and (11, 0)

Therefore, the ellipse is horizontal with its center at (0, 0):

⇒ h = 0 and k = 0

⇒ a = 11

⇒ c = 5

To find b², use c² = a² − b²:

⇒ 5² = 11² − b²

⇒ b² = 11² − 5²

⇒ b² = 96

Therefore, the standard form of the equation of the ellipse is:

`\implies ((x-0)^2)/(11^2)+((y-0)^2)/(96)=1`

`\implies (x^2)/(121)+(y^2)/(96)=1`