From simple queries to complex problems, IDNLearn.com provides reliable answers. Ask your questions and receive reliable and comprehensive answers from our dedicated community of professionals.

1. Write an equation in standard form of an ellipse that has a vertex at [tex]\((0, 6)\)[/tex], a co-vertex at [tex]\((1, 0)\)[/tex], and a center at the origin.

A. [tex]\[ x^2 + \frac{y^2}{6} = 1 \][/tex]

B. [tex]\[ \frac{x^2}{36} + \frac{y^2}{2} = 1 \][/tex]

C. [tex]\[ \frac{x^2}{36} + y^2 = 1 \][/tex]

D. [tex]\[ x^2 + \frac{y^2}{36} = 1 \][/tex]


Sagot :

To find the standard form of the equation of an ellipse that has a vertex at (0,6), a co-vertex at (1,0), and a center at the origin (0,0), follow these steps:

1. Identify the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
- The vertex (0,6) indicates that the distance from the center to a vertex along the y-axis is [tex]\(a = 6\)[/tex].
- The co-vertex (1,0) indicates that the distance from the center to a co-vertex along the x-axis is [tex]\(b = 1\)[/tex].

2. Calculate [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
- [tex]\(a^2 = 6^2 = 36\)[/tex]
- [tex]\(b^2 = 1^2 = 1\)[/tex]

3. Write the standard form of the ellipse:
- For an ellipse with a vertical major axis centered at the origin, the standard form of the equation is:
[tex]\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \][/tex]

4. Plug the values of [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex] into the equation:
- Substituting [tex]\(a^2 = 36\)[/tex] and [tex]\(b^2 = 1\)[/tex], the equation becomes:
[tex]\[ \frac{x^2}{1} + \frac{y^2}{36} = 1 \][/tex]

So, the correct equation in standard form is:
[tex]\[ \boxed{\frac{x^2}{1} + \frac{y^2}{36} = 1} \][/tex]