### introducing Focal Radi, pythagoren and standerd form for the equatoin of an ellips.

Today was a one period class and i thank abr13l for choosing me:D because last time i did a two period scribe and i did not like it :D

Today's class was skatching graphs, finding the vertex, finding the foci and writting some notes.

We started class by two questions on the board.

Write the equation in standard form and.....

a) sketch the graph

b)Find the vertex

c)Find the foci

I) X2-4y2-8=0

solution:

x2/8 -4y2 =8/8

x2/8-y2/2=1

c2=a2-b2

c2=8-2

c=sqroot 6

a=2sq2 b=sq2 c=sq6

the vertex= (0,0)

the foci=sq6

II)9x2+4y2-54x+36=0

solution:

9x2-54x 4y2+36=-36

9(x2-6x+9) +4(y2+9y+81/4)=-36+81+81

9(x-3)2 +4(y+9/2)2=126

(1/9)/(1/9) 9(x-3)2/126+(1/4)/(1/4) 4(y=9/2)/126=1

(x-3)2/126/9+(y+9/2)/126/9=1

(x-3)2/14+(y+9/2)2/63/2

the vertex =(3,4 1/2)

the foci=(3,10)

(Sorry i couldn't post the graph)

Then it was time for note taking. Today we had a fair amount of notes to take.

(I couldn't post the ellips to show you where these letters and notes had came from)

O IS THE CENTRE WITH THE COORDINATES (h,k)

A1 A2 is the majour axis its length is 2a

B1B2 is the minor axis ; its length is 2a

OA1 is the semi- majour axis; its length is a

OB1 is the semi-minor axis;its length is b

F1 and F2 are the foci of the ellips, each one is c units from the centre.

PF1 & PF2 are the focal radii

A1&A2 are called the verticies of the ellips they are the end point of the majour axis.

The Focal Radi property ( I couldn't post the ellips)

By defination PF1+PF2 is constant assum that P is at A1

therefore PF1+PF2=A1F1+A2F2

By summary A1F1=A2F2

=A2F1+A1F2

=2a

The phthagoren property: (I couldn't post the ellips)

By defination PF1 +PF2 is constant and we know that PF1+pF2= 2a

Assum P is at B1 By summary PF1=PF2

PF1=PF2=2a

2PF1=2a

PF1=a

therefore c2+b2=a2

or c2-a2=b2

By defination PF1+PF2 is constant assum that P is at A1

therefore PF1+PF2=A1F1+A2F2

By summary A1F1=A2F2

=A2F1+A1F2

=2a

The phthagoren property: (I couldn't post the ellips)

By defination PF1 +PF2 is constant and we know that PF1+pF2= 2a

Assum P is at B1 By summary PF1=PF2

PF1=PF2=2a

2PF1=2a

PF1=a

therefore c2+b2=a2

or c2-a2=b2

Horizontal Vertival orientation

(x-h)2/a+(y-k)2/b2=1 (x-h)2/b2+(y-k)2/a2=1

(I couldn't post the ellips.)

We have an assignment parabolas,circles,and ellipses. This assignment shows every thing we did in conics so far.

And the next scribe is Jessica (good luck on a two period scribe)

## 2 Comments:

I know this is a few years late, but anyone searching this page for help on this exercise needs to know the answer provided for c on problem 1 is wrong.

c^2 = b^2 + a^2 is the correct one.

@Anonymous If "c" represents the hypotenuse of a right-angled triangle then you'd be exactly right. In the example here, "c" does not represent the hypotenuse of a right-angled trigle, "a" does.

Unfortunately zaenab didn't post a diagram to illustrate her work. I wonder if you could construct the diagram she would have posted if she could have? It would be really wonderful if you could leave a link to it here too. ;-)

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