Question

if y=mx+c, and x^2/a^2 + y^2/b^2 = 1, prove
that c^2 - b^2 = (am)^2

Answer 1

substitute for y and see what happens !

Answer 2

i tried, some nonsense came, and it's too many variables turning my brain into mush. X_X

Answer 3

lol

Answer 4

lol

Answer 5

lol is RIGHT!

Answer 6

A good one....

Answer 7

!? :-/

Answer 8

I'm solving it but i think i made a mistake somewhere along the line... :/ i'll try again

Answer 9

Problem is we have two c's??

One is distance from foci to centre =a^2-b^2
and the other hanging around in the line equation...

Answer 10

Let's call the ellipse c C instead.

Answer 11

Where's the op? Which c is what c?
Not doing anything until u tell me.

Answer 12

omg i tried so hard to solve this lol i think im missing something

Answer 13

u c....:-)

Answer 14

^LOL

Answer 15

I give up! darn it. anyone else still working on this? lemme know when u figure it out

Answer 16

if y=mx+c, and x^2/a^2 + y^2/b^2 = 1,
prove c^2 - b^2 = (am)^2

Consider m=1, c=2, a=2, b=1
See http://www.wolframalpha.com/input/?i=solve+y%3D+x%2B2%3B+%28x%2F2%29%5E2%2B%28y%2F1%29%5E2%3D1

then c^2-b^2 = 4-1= 3
(am)^2= (2*1)^2= 4

So unless we have an additional constraint, this is not true for the general case.

Answer 17

Ah but which c is it?

Answer 18

I only see 1 c in the given equations. Is there another one?

Answer 19

trying to solve and prove it, you get 2 c's, one is squared, the other isn't

Answer 20

Some schools give focus distance as c^2 = a^2 -b^2..

Answer 21

OK, I'll see if that makes any sense...

Answer 22

I said to op I wouldn't do anything until she said which c was which...

Answer 23

But mia since posting....

Answer 24

If we assume c^2= a^2-b^2, then, in my test case,
c^2- b^2= a^2-2b^2= 4-2=2
(am)^2= 4
No, it does not work!