Question

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

Explanation:

Step1: Identify the type of conic section

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

Step2: Find the vertices

For the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we have $a^{2}=9$, so $a = 3$, and $b^{2}=4$, so $b = 2$.
The vertices along the $y$-axis are $(0,a)=(0,3)$ and $(0,-a)=(0, - 3)$.
The co - vertices along the $x$-axis are $(b,0)=(2,0)$ and $(-b,0)=(-2,0)$.

Step3: Plot the points and draw the ellipse

Plot the points $(0,3),(0, - 3),(2,0),(-2,0)$ on the coordinate plane. Then sketch the ellipse passing through these four points.

Answer:

Plot the points $(0,3),(0, - 3),(2,0),(-2,0)$ and draw an ellipse passing through them.