Answer
Refer to the graph below.
Work Step by Step
RECALL:
The graph of $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ where $a\gt b$, is a vertical ellipse that has:
its center at $(0, 0)$
major axis length = $2a$
minor axis length = $2b$
Before the given inequality can be graphed, there is a need to graph first the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$.
The equation $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$ can be written as $\dfrac{x^2}{3^2}+\dfrac{y^2}{4^2}=1$ therefore is has $a=4$ and $b=3$. Thus, ots graph is a vertical ellipse whose center is at $(0, 0)$ and whose major axis is 8 units long and whose minor axis is 3 units long.
Graph the ellipse. (Refer to the graph below.)
The graph of the inequality $\dfrac{x^2}{9} + \dfrac{y^2}{16^2} \le1$ includes the ellipse itself and the region inside the ellipse. Thus, shade the region inside the ellipse.
(Refer to the graph in the answer part above.)
