graph each equation.\n9) $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$

Explanation:

Step1: Identify the form of the equation

The equation $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is in the standard form of an ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1$ ($a>b>0$). Here, $a^{2}=9$, so $a = 3$; $b^{2}=4$, so $b=2$.

Step2: Find the vertices

For the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, the vertices are at $(0,\pm a)$. Substituting $a = 3$, the vertices are $(0,3)$ and $(0, - 3)$.

Step3: Find the co - vertices

The co - vertices are at $(\pm b,0)$. Substituting $b = 2$, the co - vertices are $(2,0)$ and $(-2,0)$.

Step4: Sketch the ellipse

Plot the vertices $(0,3),(0, - 3)$ and co - vertices $(2,0),(-2,0)$. Then draw a smooth curve passing through these four points to form the ellipse.

Answer:

The ellipse has vertices at $(0,3)$ and $(0, - 3)$, co - vertices at $(2,0)$ and $(-2,0)$. Sketch the ellipse passing through these four points.