Question

graph the equation: $\frac{x^{2}}{4}+\frac{y^{2}}{16}=1$

Explanation:

Step1: Identify the type of conic section

The equation $\frac{x^{2}}{4}+\frac{y^{2}}{16}=1$ is in the form of $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1$ ($a>b>0$), which represents an ellipse centered at the origin $(0,0)$.

Step2: Find the vertices

For the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we have $a^{2}=16$ and $b^{2} = 4$. So $a = 4$ and $b=2$.
The vertices of the ellipse (along the $y$ - axis since $a>b$) are $(0,a)$ and $(0, - a)$, i.e., $(0,4)$ and $(0,-4)$.
The co - vertices (along the $x$ - axis) are $(b,0)$ and $(-b,0)$, i.e., $(2,0)$ and $(-2,0)$.

Step3: Sketch the ellipse

Plot the center at $(0,0)$, vertices at $(0,4)$ and $(0,-4)$, co - vertices at $(2,0)$ and $(-2,0)$. Then draw the smooth curve of the ellipse passing through these four points.

Answer:

Sketch an ellipse centered at the origin with vertices $(0,4)$ and $(0,-4)$ and co - vertices $(2,0)$ and $(-2,0)$.