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 ~

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

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

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

$\qquad \sf  \dashrightarrow \: b {}^{2} = 96$

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

Answer:

$\dfrac{x^2}{121}+\dfrac{y^2}{96}=1$

Step-by-step explanation:

General equation of an ellipse:

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$

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:

$\implies \dfrac{(x-0)^2}{11^2}+\dfrac{(y-0)^2}{96}=1$

$\implies \dfrac{x^2}{121}+\dfrac{y^2}{96}=1$