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.
y^2=x ⇔ x=y^2
We notice that the variable raised to the power of 2 is y. Therefore, we recognize this as a horizontal parabola. We will start by writing the given equation in the form x= a(y- k)^2+ h.
x=y^2 ⇔ x= 1(y- 0)^2 + 0
We can see that a= 1, h= 0, and that k= 0. Next, we can use this information to highlight some important characteristics of the parabola.
x=a(y-k)^2+h
x=1(y-0)^2+ 0
Direction of Opening
Right if a>0, Left if a<0
Right ( 1>0)
Vertex
( h, k)
( 0, 0)
Axis of Symmetry
y= k
y= 0
Focus
( h+1/4 a, k)
( 0+1/4( 1), 0) ⇕ (1/4,0 )
Directrix
x= h-1/4 a
x= 0-1/4( 1) ⇕ x=- 1/4
Length of Latus Rectum
|1/a| units
|1/1| units ⇕ 1 units
We can use the above information to draw the graph.
Now that we have the boundary line, we need to determine which region to shade. To do so, we will use (2,0) 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 (2,0). Because we have a strict inequality, the boundary line will be dashed.
Inequality (II)
We can write the boundary curve by replacing the less than sign with an equals sign.
x^2-4y^2= 16
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. Next, let's review the main characteristics of this type of hyperbola.
Using this information, we can identify that the vertices are (± 4 ,0). Let's substitute a=4 and b=2 into the formula for the asymptotes and obtain their equations.
y=± b/ax ⇒ y=± 2/4x=± 1/2x
The asymptotes are y=± 12x. Now, let's calculate c, the absolute value of the nonzero coordinate of the foci. To do so, we will substitute a=4 and b=2 into c^2=a^2+b^2.
The foci of the hyperbola are (± sqrt(20),0). To graph the function, let's summarize all of the information that we have found.
Equation
x^2/4^2-y^2/2^2=1
Transverse axis
Horizontal
Vertices
(± 4,0)
Foci
(± sqrt(20),0)
Asymptotes
y=± 1/2x
Let's now draw the hyperbola. Because the inequality is strict, the boundary curve will be dashed.
Now that we have the boundary curve, we need to determine which region to shade. This time we will use (0,2) as a test point. Let's determine if the point satisfies the inequality or not.
