We can write the boundary curve for Inequality (I) by replacing the less than sign with an equals sign. Then, we can identify which conic section it is.
x^2+y^2=49
Let's recall the standard equation of a circle.
(x- h)^2+(y- k)^2= r^2
Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius.
x^2+y^2=49 [0.3em] ⇕ [0.3em] (x- 0)^2+(y- 0)^2= 7^2
The center of the circle is the point ( 0, 0), and its radius is 7. Let's draw its graph.
Now that we have the boundary curve, we need to determine which region to shade. To do so, we will use (0,0) as a test point. If the point satisfies the inequality, we will shade the region that contains the point. If not, we will shade the opposite region.
Since the substitution produced a true statement, we will shade the region that contains the point (0,0). Because we have a strict inequality, the boundary curve will be dashed.
Inequality (II)
We can write the boundary curve by replacing the greater than or equal to sign with an equals sign.
16x^2-9y^2=144
To identify which conic section it is we need to rewrite the equation.
Let's recall the equation of hyperbolas centered at the origin.
ccc
Horizontal & & Vertical
Hyperbola & & Hyperbola
x^2/a^2-y^2/b^2=1 & & y^2/a^2-x^2/b^2=1
From the above formula, we can see that our equation represents a horizontal hyperbola. Let's review the main characteristics of this type of hyperbola.
Using this information, we can identify that the vertices are (± 3 ,0). Let's substitute a=3 and b=4 into the formula for the asymptotes and obtain their equations.
y=± b/ax ⇒ y=± 4/3x
The asymptotes are y=± 43x. Now, let's calculate c, the absolute value of the nonzero coordinate of the foci. To do so, we will substitute a=3 and b=4 into c^2=a^2+b^2.
The foci of the hyperbola are (± 5,0). To graph the function, let's summarize all of the information that we have found.
Equation
x^2/3^2-y^2/4^2=1
Transverse axis
Horizontal
Vertices
(± 3,0)
Foci
(± 5, 0)
Asymptotes
y=± 4/3x
Let's now draw the hyperbola on the same set of axis as we drew the circle. Because the inequality is non-strict, the boundary curve will be solid.
Now that we have the boundary curve, we need to determine which region to shade. Again, we will use (0,0) as a test point. Let's determine if the point satisfies the inequality or not.
