Given :
∙ Fixed point (F)
∙ Fixed line or directrix (l)
∙ Eccentricity (e = 1)
∙ Moving point P(x, y)
Construction :
∙ Plot the fixed point F and draw the fixed line ‘l’.
∙ Drop a perpendicular (FZ) from F to l.
∙ Draw a perpendicular bisector to FZ and treat it as y-axis.
∙ Let V(0, 0) be the origin.
∙ Drop a perpendicular (PM) from P to l.
∙ The known points are F(a, 0), Z(-a, 0) and M is (-a, y).
By the definition of a conic,
FP/PM = e = 1
FP/PM = 1
FP = PM
Square both sides.
FP2 = PM2
(x - a)2 + (y - 0)2 = (x + a)2 + (y - y)2
x2 - 2ax + a2 + y2 = x2 + 2ax + a2
Subtract x2 and a2 from each side.
- 2ax + y2 = 2ax
Add 2ax to each side.
y2 = 4ax
This is the standard equation of the parabola with vertex at origin.
Graph of y2 = 4ax :
If the vertex is other then origin, say (h, k), then
(y - k)2 = 4a(x - h)
1. Open leftward : y2 = -4ax [a > 0]
2. Open upward : x2 = 4ay [a > 0]
3. Open downward : x2 = -4ay [a > 0]
Use the information provided to write the standard equation of each parabola.
Problem 1 :
Vertex at origin, Focus (0, 1)
Solution :
Plot the vertex (0, 0) and focus (0, 1) on the xy-plane.
The parabola is open up with vertex at origin.
Standard equation of a parabola that opens up with vertex at origin :
x2 = 4ay
Distance between the vertex and focus is 1 unit.
That is, a = 1.
x2 = 4(1)y
x2 = 4y
Problem 2 :
Vertex at origin, Focus (2, 0)
Solution :
Plot the vertex (0, 0) and focus (2, 0) on the xy-plane.
The parabola is open to the right with vertex at origin.
Standard equation of a parabola that opens right with vertex at origin :
y2 = 4ax
Distance between the vertex and focus is 2 units.
That is, a = 2.
y2 = 4(2)x
y2 = 8x
Problem 3 :
Vertex at (1, 2), Focus (1, -1)
Solution :
Plot the vertex (1, 2) and focus (1, -1) on the xy-plane.
The parabola is open down with vertex at (1, 2).
Standard equation of a parabola that opens down with vertex at (h, k) :
(x - k)2 = -4a(y - h)
Vertex (h, k) = (1, 2).
(x - 1)2 = -4a(y - 2)
Distance between the vertex and focus is 3 units.
That is, a = 3.
(x - 1)2 = -4(3)(y - 2)
(x - 1)2 = -12(y - 2)
Problem 4 :
Vertex at (2, -1), Focus (-1, -1)
Solution :
Plot the vertex (2, -1) and focus (-1, -1) on the xy-plane.
The parabola is open to the left with vertex at (2, -1).
Standard equation of a parabola that opens left with vertex at (h, k) :
(y - k)2 = -4a(x - h)
Vertex (h, k) = (2, -1).
(y + 1)2 = -4a(x - 2)
Distance between the vertex and focus is 3 units.
That is, a = 3.
(y + 1)2 = -4(3)(x - 2)
(y + 1)2 = -12(x - 2)
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