Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Write an equation for an ellipse centered at the origin, which has foci at
(+-12,0) and vertices at
(+-13,0).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm 12,0)$ and vertices at $( \pm 13,0)$ .

View step-by-step help

View step-by-step help

Write equations of ellipses in standard form using properties

Full solution

Q. Write an equation for an ellipse centered at the origin, which has foci at

( \pm 12,0)

and vertices at

( \pm 13,0)

.

Given Foci and Vertices: We are given the foci and vertices of an ellipse centered at the origin. The distance from the center to a focus is denoted by $c$ , and the distance from the center to a vertex is denoted by $a$ . Since the ellipse is centered at the origin, we can directly read these values from the coordinates of the foci and vertices. $\newline$ Foci: $(\pm 12, 0)$ implies $c = 12$ $\newline$ Vertices: $(\pm 13, 0)$ implies $a = 13$
Calculating b: The relationship between $a$ , $b$ , and $c$ for an ellipse is given by the equation $c^2 = a^2 - b^2$ , where $b$ is the distance from the center to the co-vertex. We need to find the value of $b$ using the values of $a$ and $c$ .
Calculate $b$ using the relationship $c^2 = a^2 - b^2$ .
$b$ $0$
$b$ $1$
$b$ $2$
$b$ $3$
Finding the Standard Form Equation: Now that we have $b^2$ , we can find $b$ by taking the square root of $b^2$ . $\newline$ $b = \sqrt{b^2}$ $\newline$ $b = \sqrt{25}$ $\newline$ $b = 5$
Finding the Standard Form Equation: Now that we have $b^2$ , we can find $b$ by taking the square root of $b^2$ . $\newline$ $b = \sqrt{b^2}$ $\newline$ $b = \sqrt{25}$ $\newline$ $b = 5$ With the values of $a$ and $b$ , we can write the standard form equation of the ellipse. Since the ellipse is horizontal (the vertices are on the x-axis), the standard form of the equation is $\left(\frac{x^2}{a^2}\right) + \left(\frac{y^2}{b^2}\right) = 1$ . $\newline$ Substitute $a = 13$ and $b = 5$ into the equation. $\newline$ $\left(\frac{x^2}{13^2}\right) + \left(\frac{y^2}{5^2}\right) = 1$ $\newline$ $\left(\frac{x^2}{169}\right) + \left(\frac{y^2}{25}\right) = 1$

More problems from Write equations of ellipses in standard form using properties

Question

Write the equation in standard form for the ellipse with center at the origin, vertex

(-10,0)

, and co-vertex

(0,3)

Posted 1 month ago

Question

Write the equation in standard form for the ellipse

7x^2 + y^2 = 21

\newline

Posted 2 months ago

Question

What is the center of the ellipse

x^2 + 2y^2 - 16 = 0

\newline

Write your answer in simplified, rationalized form.

\newline

(\_\_\_\_\_,\_\_\_\_\_)

Posted 1 month ago

Question

What is the length of the major axis of the ellipse

\frac{x^2}{9} + \frac{y^2}{2} = 1

\newline

Write your answer in simplified, rationalized form.

\newline

\_\_\_\_\_

Posted 1 month ago

Question

What is the center of the ellipse

\left(\frac{(x - 5)^2}{9}\right) + \left(\frac{(y - 2)^2}{54}\right) = 1

\newline

Write your answer in simplified, rationalized form.

\newline

(________ , _______)

Posted 2 months ago

Question

The equation of an ellipse is given below.

\newline

\frac{(x+15)^{2}}{676}+\frac{(y-4)^{2}}{100}=1

\newline

What are the foci of this ellipse?

\newline

1

\newline

(-15+\sqrt{24}, 4)

(-15-\sqrt{24}, 4)

\newline

(-15,4+\sqrt{24})

(-15,4-\sqrt{24})

\newline

(-39,4)

(9,4)

\newline

(-15,28)

(-15,-20)

Posted 3 months ago

Question

The equation of an ellipse is given below.

\newline

\frac{(x-5)^{2}}{3}+\frac{(y-7)^{2}}{6}=1

\newline

What are the foci of this ellipse?

\newline

1

\newline

(5,7+\sqrt{3})

(5,7-\sqrt{3})

\newline

(-5,-7+\sqrt{3})

(-5,-7-\sqrt{3})

\newline

(-5+\sqrt{3},-7)

(-5-\sqrt{3},-7)

\newline

(5+\sqrt{3}, 7)

(5-\sqrt{3}, 7)

Posted 3 months ago

Question

The equation of an ellipse is given below.

\newline

\frac{x^{2}}{46}+\frac{(y+8)^{2}}{26}=1

\newline

What are the foci of this ellipse?

\newline

1

\newline

(0+\sqrt{20}, 8)

(0-\sqrt{20}, 8)

\newline

(0,8+\sqrt{20})

(0,8-\sqrt{20})

\newline

(0,-8+\sqrt{20})

(0,-8-\sqrt{20})

\newline

(0+\sqrt{20},-8)

(0-\sqrt{20},-8)

Posted 3 months ago

Question

Write an equation for an ellipse centered at the origin, which has foci at

(0, \pm 6)

and vertices at

(0, \pm \sqrt{37})

Posted 3 months ago

Question

Answer the questions below about the quadratic function.

\newline

f(x)=x^{2}+6 x+13

\newline

Does the function have a minimum or maximum value?

\newline

\newline

\newline

Where does the minimum or maximum value occur?

\newline

x=

\newline

\square

\newline

What is the function's minimum or maximum value?

Posted 5 hours ago