Question

Identify the vertices and the foci of the ellipse.

(x^2/18)+(y^2/8)=1

Answer 1

`\textsf{Vertices:} \quad \left(-3√(2),0\right)\;\;\textsf{and}\;\;\left(3√(2),0\right)`

`\textsf{Foci:} \quad \left(-√(10),0\right)\;\;\textsf{and}\;\;\left(√(10),0\right)`

Explanation:

Given equation of the ellipse:

`(x^2)/(18)+(y^2)/(8)=1`

As the denominator of the x² term is larger than the denominator of the y² term, it indicates that the ellipse has a horizontal major axis, making it a horizontal ellipse.

The general equation for a horizontal ellipse is:

`\large\boxed{((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1}`

where:

• Center = (h, k)
• Vertices = (h±a, k)
• Co-vertices = (h, k±b)
• Foci = (h±c, k) where c² = a² - b²

For the given equation, the values of h, k, a² and b² are:

• h = 0
• k = 0
• a² = 18
• b² = 8

To find the coordinates of the vertices, first find the value of a:

`\begin{aligned}a^2&=18\\√(a^2)&=√(18)\\a&=√(18)\end{aligned}`

Now, substitute the values of h, a and k into the vertices formula:

`\begin{aligned}\sf Vertices&=(h\pm a, k)\\&=(0 \pm √(18), 0)\\&=(\pm √(18),0)\\&=(\pm √(3^2 \cdot 2),0)\\&=(\pm √(3^2)√(2),0)\\&=(\pm 3√(2),0)\end{aligned}`

Therefore, the vertices of the ellipse are:

`\left(-3√(2),0\right)\;\;\textsf{and}\;\;\left(3√(2),0\right)`

To find the coordinates of the foci, first find the value of c by substituting the values of a² and b² into c² = a² - b²:

`\begin{aligned}c^2&=a^2-b^2\\c&=\pm√(a^2-b^2)\\c&=\pm√(18-8)\\c&=\pm√(10)\end{aligned}`

Substitute the found value of c and the values of h and k into the foci formula:

`\begin{aligned}\sf Foci&=(h\pm c, k)\\&=(0 \pm √(10), 0)\\&=(\pm √(10),0)\end{aligned}`

Therefore, the foci of the ellipse are:

`\left(-√(10),0\right)\;\;\textsf{and}\;\;\left(√(10),0\right)`

Answer 2

`\sf \textsf{ Vertices of ellipse} :(\pm 3√(2),0)`

`\sf \textsf{Foci of ellipse} : ( \pm √(10),0 )`

Explanation:

Definition:

The vertices of an ellipse are the two points that are the farthest away from the center of the ellipse.

The foci of an ellipse are two points that are located on its major axis, and are the same distance away from the center of the ellipse.

Formula:

The vertices and foci of an ellipse can be found using the following formulas:

When a is greater than b.

Vertices: (h ± a , k )

Foci: (h ± c, k )

where:

• h and k are the center of the ellipse
• a is the semi-major axis
• c is the semi-minor axis

Solution:

In order to find the vertices and foci of the ellipse
`\sf (x^2)/(18)+(y^2)/(8)=1`, we first need to rewrite the equation in standard form. The standard equation of the ellipse is given by:

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

Comparing with this equation, we get

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

`\sf a^2 = 18`

`\sf a =3√(2)`

`\sf b^2 = 8`

`\sf b =√(18)= 2√(2)`

Center: (0, 0)

Now, Find the vertices:

`\sf Vertices: (0 \pm 3√(2),0)`

`\sf Vertices: (\pm 3√(2),0)`

Find the foci:

`\sf Foci: (0\pm c, 0)`

The linear eccentricity (focal distance) (c) can be calculated by the following formula:

`\sf c^2 = a^2 - b^2`

`\sf c^2 =( 3√(2))^2 - (2√(2))^2`

`\sf c^2 = 18 - 8`

`\sf c^2 = 10`

`\sf c =√(10)`

So,

`\sf Foci: ( \pm √(10),0 )`

Therefore,

`\sf \textsf{ Vertices of ellipse} :(\pm 3√(2),0)`

`\sf \textsf{Foci of ellipse} : ( \pm √(10),0 )`