Find the standard form of the equation of the ellipse with the given characteristics and center at the origin.
A) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 25 } = 1$
B) $\frac { x } { 25 } + \frac { y } { 49 } = 1$
C) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 49 } = 1$
D) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 49 } = 1$
E) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 49 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics and center at the origin.  
A) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 25 } = 1$ 
B) $\frac { x } { 25 } + \frac { y } { 49 } = 1$ 
C) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 49 } = 1$ 
D) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 49 } = 1$ 
E) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 49 } = 1$

Quizplus · Accepted Answer

We can determine the standard form of the equation of an ellipse centered at the origin by using the formula:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.
From the given characteristics, we can observe that the semi-major axis has a length of 7 and the semi-minor axis has a length of 5. Therefore, we have:
$$\frac{x^2}{7^2}+\frac{y^2}{5^2}=1$$
Simplifying this equation results in:
$$\frac{x^2}{49}+\frac{y^2}{25}=1$$
which is in the desired standard form. Therefore, the answer is choice C.

Find the Standard Form of the Equation of the Ellipse x225+y225=1\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 25 } = 125x2​+25y2​=1

Find the Standard Form of the Equation of the Ellipse $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 25 } = 1$