In practical terms, it’s often easier to recognize parabolas in three dimensions. For example, think of large parabolic satellite dishes, or the clear plastic parabolic microphones you see on the sidelines of football games. Both of these direct waves (radio, sound, etc. ) toward a single point—the focal point (or focus).
Although they’re (x,y){\displaystyle (x,y)} coordinates, you’ll see the vertex coordinates represented by (h,k){\displaystyle (h,k)} in parabola equations and formulas.
For our needs, it’s also important that the vertex is always exactly halfway between the focus and the directrix along the axis of symmetry.
The distance between the vertex and the directrix (at the axis of symmetry) is always exactly the same as that between the vertex and the focus.
If you draw a straight line from the focus to any point along the curve of the parabola, and then draw a straight line from that point to intersect at a right angle with the directrix, you’ll find that both of those lines are always equal in length.
If you have a graph of the parabola, it’s easy to tell which equation to use. But what if you’re only given the parabola in equation form? Here’s the trick to use: If the x{\displaystyle x} component is squared in the parabola’s equation—for example y=x2{\displaystyle y=x^{2}}—convert it into the form y=a(x−h2)+k{\displaystyle y=a(x-h^{2})+k}. If the y{\displaystyle y} component is squared (like in x=y2{\displaystyle x=y^{2}}), use x=a(y−k)2+h{\displaystyle x=a(y-k)^{2}+h}. In both equations, (h,k){\displaystyle (h,k)} represent the coordinates of the parabola’s vertex, and a{\displaystyle a} represents its slope.
Use (h,k+1/(4a)){\displaystyle (h,k+1/(4a))} for a “regular” parabola in form y=a(x−h2)+k{\displaystyle y=a(x-h^{2})+k}. Use (h+1/(4a),k){\displaystyle (h+1/(4a),k)} for a “sideways” parabola in form x=a(y−k)2+h{\displaystyle x=a(y-k)^{2}+h}.
y=x2{\displaystyle y=x^{2}} → y=1(x−0)2+0{\displaystyle y=1(x-0)^{2}+0} Also note that, because a{\displaystyle a} is positive, the parabola opens upward.
That means the focus equation is (h,k+1/(4a)){\displaystyle (h,k+1/(4a))} → (0,0+1/(4a)){\displaystyle (0,0+1/(4a))}
a=1{\displaystyle a=1} in y=1(x−0)2+0{\displaystyle y=1(x-0)^{2}+0} So, the focus equation is (0,0+1/(4(1))){\displaystyle (0,0+1/(4(1)))} Therefore, the focus of y=x2{\displaystyle y=x^{2}} is at (0,1/4){\displaystyle (0,1/4)}
x=y2{\displaystyle x=y^{2}} → x=1(y−0)2+0{\displaystyle x=1(y-0)^{2}+0} Vertex (h,k)=(0,0){\displaystyle (h,k)=(0,0)}, a=1{\displaystyle a=1} (h+1/(4a),k){\displaystyle (h+1/(4a),k)} → (0+1/(4(1)),0){\displaystyle (0+1/(4(1)),0)} → (1/4,0){\displaystyle (1/4,0)} Answer: The focus is located at (1/4,0){\displaystyle (1/4,0)}.
Put (y−3)2=8(x−5){\displaystyle (y-3)^{2}=8(x-5)} in x=a(y−k)2+h{\displaystyle x=a(y-k)^{2}+h} form: (x−5)=1/8(y−3)2){\displaystyle (x-5)=1/8(y-3)^{2})} → x=1/8(y−3)2+5{\displaystyle x=1/8(y-3)^{2}+5} Vertex (h,k)=(5,3){\displaystyle (h,k)=(5,3)}, a=1/8{\displaystyle a=1/8} (h+1/(4a),k){\displaystyle (h+1/(4a),k)} → (5+1/(4(1/8)),3){\displaystyle (5+1/(4(1/8)),3)} (5+1/(1/2),3){\displaystyle (5+1/(1/2),3)} →(5+2,3){\displaystyle (5+2,3)} Answer: The focus is located at (7,3){\displaystyle (7,3)}.
Put (x+3)2=−20(y−1){\displaystyle (x+3)^{2}=-20(y-1)} in the proper form: (y−1)=−1/20(x+3)2{\displaystyle (y-1)=-1/20(x+3)^{2}} → y=−1/20(x+3)2+1{\displaystyle y=-1/20(x+3)^{2}+1} Vertex (h,k)=(−3,1){\displaystyle (h,k)=(-3,1)}, a=−1/20{\displaystyle a=-1/20} (h,k+1/(4a)){\displaystyle (h,k+1/(4a))} → (−3,1+1/(4(−1/20))){\displaystyle (-3,1+1/(4(-1/20)))} (−3,1+1/(−1/5){\displaystyle (-3,1+1/(-1/5)} → (−3,1+(−5)){\displaystyle (-3,1+(-5))} Answer: The focus is located at (−3,−4){\displaystyle (-3,-4)}
Put (x−7)2=3(y−4){\displaystyle (x-7)^{2}=3(y-4)} in the proper form: (y−4)=1/3(x−7)2{\displaystyle (y-4)=1/3(x-7)^{2}} → y=1/3(x−7)2+4{\displaystyle y=1/3(x-7)^{2}+4} Vertex (h,k)=(7,4){\displaystyle (h,k)=(7,4)}, a=1/3{\displaystyle a=1/3} (h,k+1/(4a)){\displaystyle (h,k+1/(4a))} → (7,4+1/(4(1/3))){\displaystyle (7,4+1/(4(1/3)))} (7,4+1/(4/3)){\displaystyle (7,4+1/(4/3))} → (7,4+(3/4)){\displaystyle (7,4+(3/4))} → (7,(16/4)+(3/4)){\displaystyle (7,(16/4)+(3/4))} Answer: the focus is located at (7,19/4){\displaystyle (7,19/4)}
