MAT01B1: Parabolas and Hyperbolas

Dr Craig

30 October 2018

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I acraig@uj.ac.za

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I Today

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I D1 LAB 308

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Parabolas: a parabola is the set of points

in a plane that are equidistant from a fixed

point F and a fixed line. The point F is

called the focus and the line is called the

directrix.

Parabolas

The vertex is the point of the parabola that

is on the line perpendicular to the directrix

that goes through F .

A parabola with focus (0, p) and directrix

y = −p has equation

x2 = 4py

If we interchange x and y we get

y2 = 4px, focus (p, 0), directrix x = −p

Parabola examples

Find the focus and directrix of the parabola

y2 + 10x = 0.

Find the vertex, focus and directrix of the

the parabola 2x = −y2.

A shifted parabola:

Sketch y2 + 2y − x = 0.

Hyperbolas

A hyperbola is the set of all points in a

plane the difference of whose distances from

two fixed points F1 and F2 is a constant.

|PF1| − |PF2| = ±2a

We can show that if the foci of a hyperbola

are on the x-axis at (±c, 0) and we have

|PF1| − |PF2| = ±2a, then the equation is

x2

a2− y2

b2= 1 (c2 = a2 + b2)

Equations of a hyperbola (1)

The hyperbola

x2

a2− y2

b2= 1

has foci (±c, 0), where c2 = a2 + b2,

vertices (±a, 0), and asymptotes

y = ±(b/a)x.

Equations of a hyperbola (2)

We can also have the foci of a hyperbola on

the y-axis.

The hyperbola

y2

a2− x2

b2= 1

has foci (0,±c), where c2 = a2 + b2,

vertices (0,±a), and asymptotes

y = ±(a/b)x.

Important hyperbolic facts

I a hyperbola where the x2 has a positive

coefficient looks like an x

I the value of a is used to find the

coordinates of the vertices

I the a2 is always below the term with the

positive coefficient

I there is no required relationship between

a and b

Hyperbola example 1

Find the foci and asymptotes of the

hyperbola 9x2 − 16y2 = 144 and sketch its

graph.

Hyperbola example 2

Find the foci and equation of the hyperbola

with vertices (0,±1) and asymptote y = 2x.

Sketch of example 2: y2 − 4x2 = 1

Yet another shifted conic:

Consider the curve

4x2 − y2 − 24x− 4y + 16 = 0.

We can tell immediately that this is a

hyperbola. However we must first do some

algebra before deciding which shape it will

be.

4x2 − y2 − 24x− 4y + 16 = 0

(x− 3)2

4− (y + 2)2

16= 1

Shifted example 2

Sketch and find the foci of the conic

9x2 − 4y2 − 72x + 8y + 176 = 0.

Online sketcher

Use the following website to help familiarise

yourself with conic sections:

https://www.desmos.com/calculator/vgfqbejegx

To change it to the equation of an ellipse

just click the function box and change the

minus to a plus.

You can also click the 3 horizontal lines in

the top left corner to sketch other types of

curves.

A shifted ellipse: consider

(x− 2)2

3+

(y + 1)2

2= 1.

This will be the same shape as the ellipse

x2

3+

y2

2= 1

but shifted 2 units to the right and 1 unit

down. Also, we have a =√3 and b =

√2.

Now, find the centre, vertices and foci and

then sketch 4x2 + y2 − 8x + 4y + 4 = 0.

(x− 2)2

3+

(y + 1)2

2= 1

(x− 1)2 +(y + 2)2

4= 1