# Parabola as Locus

The locus of a point that is equidistant from a fixed point and a fixed line is always a parabola. The fixed point is called the focus and the fixed line is called the directrix.

The locus of point P(x, y) moving so that it is equidistant from the point (0, a) and the line y=-a is a parabola with equation:

${ x }^{ 2 }=4ay$

The parabola ${ x }^{ 2 }=4ay$ has:

• Focus at (0, a)
• Directrix with equation y=-a
• Vertex at (0, 0)
• Axis with equation x=0
• Focal Length the distance from the vertex to the focus with length a
• Latus rectum that is a horizontal focal chord with length 4a

### Examples:

1. Find the equation of the parabola whose focus has coordinates (0, 2) and whose directrix has equation y=-2.

The focus has coordinates in the form (0, a) and the directrix has equation in the form y=-a, where a=2. Hence:

${ x }^{ 2 }=4(2)y \\ { x }^{ 2 }=8y$

2. a) Find the coordinates of the focus and the equation of the directrix of the parabola ${ x }^{ 2 }=20y$.

b) Find the points on the parabola at the endpoints of the latus rectum and find its length.

3. Find the coordinates of the vertex, the coordinates of the focus and the equation of the directrix of the parabola ${ x }^{ 2 }=-12y$.

The focus is in the form (0, -a) where a=3. So the focus is (0, -3).

The directrix is in the form y=a where a=3. So the directrix is y=3.

4. Find the equation of the focal chord to the parabola ${ x }^{ 2 }=4y$ that passes through (-4, 4).

The locus of point P(x, y) moving so that it is equidistant from the point P(x, y) moving so that it is equidistant from the point (a, 0) and the line x=-a is a parabola with equation:

${ y }^{ 2 }=4ax$

The parabola ${ y }^{ 2 }=4ax$ has:

• Focus at (a, 0)
• Directrix with equation x=-a
• Vertex at (0, 0)
• Axis with equation y=0
• Focal Length the distance from the vertex to the focus with length a
• Latus rectum that is a vertical focal chord with length 4a

### Examples:

1. Find the coordinates of the focus and the equation of the directrix of the parabola ${ y }^{ 2 }=32x$.

2. Find the coordinates of the focus and the equation of the directrix of the parabola ${ y }^{ 2 }=-2x$.

The focus is in the form (-a, 0) where $a=\cfrac{1}{2}$. So the focus is $(\cfrac{-1}{2}, 0)$.

The directrix is in the form x=a where $a=\cfrac{1}{2}$. So the directrix is $x=\cfrac{1}{2}$.

General Parabola (Concave Up or Down)

The concave upwards parabola with vertex (h, k) and focal length a has equation:

${ \left( x-h \right) }^{ 2 }=4a(y-k)$

The parabola ${ \left( x-h \right) }^{ 2 }=4a(y-k)$ has:

• Axis parallel to the y-axis
• Vertex at (h, k)
• Focus at (h, k+a)
• Directrix with equation y=k-a

The concave downwards parabola with vertex (h, k) and focal length a has equation:

${ \left( x-h \right) }^{ 2 }=-4a( y-k)$

The parabola ${ \left( x-h \right) }^{ 2 }=-4a( y-k)$ has:

• Axis parallel to the y-axis
• Vertex at (h, k)
• Focus at (h, k-a)
• Directrix with equation y=k+a

### Examples:

1. Find the equation of the parabola with focus (2, 3) and directrix with equation y=-7.

2. Find the coordinates of the vertex and the focus, and the equation of the directrix, of the parabola with equation ${ x }^{ 2 }-8x+8y-16=0$.

General Parabola (Side)

The parabola with vertex (h, k) and focal length a that turns to the right has equation:

${ \left( y-k \right) }^{ 2 }=4a{ (x-h) }$

The parabola ${ \left( y-k \right) }^{ 2 }=4a{ (x-h) }$ has

• Axis parallel to the x-axis
• Vertex at (h, k)
• Focus at (h+a, k)
• Directrix with equation x=h-a

The parabola with vertex (h, k) and focal length a that turns to the left has equation:

${ \left( y-k \right) }^{ 2 }=-4a{ (x-h) }$

The parabola ${ \left( y-k \right) }^{ 2 }=4a{ (x-h) }$ has

• Axis parallel to the x-axis
• Vertex at (h, k)
• Focus at (h-a, k)
• Directrix with equation x=h+a

### Examples:

1. Find the equation of the parabola with focus (2, 1) and directrix x=3.

2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola ${ y }^{ 2 }+12y-4x-8=0$.

## Tangent and Normals

The normal to the curve is perpendicular to its tangent at that point.
That is, $m_1 m_2=-1$for perpendicular lines.

### Examples

1. Find the gradient of the tangent to the parabola $x^2=8y$ at the point (4, 2).

${ x }^{ 2 }=8y\\ y=\cfrac { { x }^{ 2 } }{ 8 } \\ \cfrac { dy }{ dx } =\cfrac { 2x }{ 8 } \\ \cfrac { dy }{ dx } =\cfrac { x }{ 4 } \\ \\ \therefore \quad At\quad (4,2),\quad \cfrac { dy }{ dx } =\cfrac { 4 }{ 4 } =1$

So the gradient of the tangent at (4, 2) is 1.

2. Find the equation of the normal to the parabola $x^2=4y$ at the point (-8, 16).