Chapter 10.3, Problem 14AYU

In problems 13-16, the graph of an ellipse is given. Match each graph to its equation.

(A) $\frac{x^2}{4}+y^2=1$

(B) $x^2+\frac{y^2}{4}=1$

(C) $\frac{x^2}{16}+\frac{y^2}{4}=1$

(D) $\frac{x^2}{4}+\frac{y^2}{16}=1$

To determine

The equation from the following that matches with the graph.

A. $\frac{x^2}{4}+y²=1$

B. $x^2+\frac{y^2}{4}$

C. $\frac{x^2}{16}+\frac{y^2}{4}=1$

D. $\frac{x^2}{4}+\frac{y^2}{16}=1$

Answer to Problem 14AYU

Solution:

D. $\frac{x^2}{4}+\frac{y^2}{16}=1$

Explanation of Solution

Given:

Formula Used:

An equation of the ellipse with center at $\left(0,0\right)$ and vertices $\left(0, -a\right) \text{and} \left(0, a\right)$ is given by $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$.

$b$ is obtained by finding the $x\text{-intercept}$ from the graph.

Calculation:

From the graph we see that,

Vertex $=\left(0,a\right)=\left(0,4\right)$ and $\left(0, -a\right)=\left(0, -4\right)$

Center $= \left(0,0\right)$

Major axis is along $y\text{-axis}$.

$x\text{-intercepts}$ are $\left(0,b\right)=\left(0, -2\right)$ and $\left(0,-b\right)=\left(0, 2\right)$.

Equation of the ellipse is given by,

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Substituting the value for $a$ and $b$, we get,

$\frac{x^2}{2^2}+\frac{y^2}{4^2}=1$

$\frac{x^2}{4}+\frac{y^2}{16}=1$

Hence, the graph matches with option $\left(D\right)$.