**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.

FP^{2} = PM^{2}

(x - a)^{2} + (y - 0)^{2} = (x + a)^{2} + (y - y)^{2}

x^{2} - 2ax + a^{2 }+ y^{2} = x^{2} + 2ax + a^{2}

Subtract x^{2} and a^{2} from each side.

- 2ax^{ }+ y^{2} = 2ax

Add 2ax to each side.

y^{2} = 4ax

This is the standard equation of the parabola with vertex at origin.

**Graph of y ^{2} = 4ax : **

If the vertex is other then origin, say (h, k), then

(y - k)^{2} = 4a(x - h)

1. Open leftward : y^{2} = -4ax [a > 0]

2. Open upward : x^{2} = 4ay [a > 0]

3. Open downward : x^{2} = -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 :

x^{2} = 4ay

Distance between the vertex and focus is 1 unit.

That is, a = 1.

x^{2} = 4(1)y

x^{2} = 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 :

y^{2} = 4ax

Distance between the vertex and focus is 2 units.

That is, a = 2.

y^{2} = 4(2)x

y^{2} = 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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