Question

Conic sections

Answer 1

Find the equation of a parabola the extremities of whose latus rectum is (1,2) and (1,-4)

Answer 2

Hmm, I guess those points will satisfy the general equation of parabola?

Answer 3

But what is the general equation , how would we know the shape

Answer 4

We can find the length of latus rectum from here.

Answer 5

yes

Answer 6

6

Answer 7

So, 4a = 6... right?

Answer 8

yes

Answer 9

a= 6/4

Answer 10

hmm!

so... a = 6/4 = 3/2

now.. put this in at1^2 , 2at1 and at2^2 ,2at2

Answer 11

hmm .. bad idea, I guess :/

Answer 12

we can also find the focus

Answer 13

general form of parabola is y^2 = 4ax, right?

Answer 14

focus co-ordinates

Answer 15

oh yeah, better!

Answer 16

but after that

Answer 17

hmm,,.. wait I think I got it

Answer 18

how?

Answer 19

hmm.. just one minute.

Answer 20

ok

Answer 21

Given (1,2) and (1,-4) are the extremities of the focal chord of a given parabola.

Let the ends be : \((a {t_1}^2 , 2at_1 ) \ \& \ (a {t_2}^2 , 2at_2) \)

So, \(a(t_1)^2 = 1\) ; \(a = \cfrac{1}{(t_1)^2} \)

and \(2a t_1 = 2 \) => \(2 \times \cfrac{1}{(t_1)^2} = 2\)

Answer 22

Nope! doesn't work.

Let us try this -

We've focus = (1,-1)

and focus = (h,k+a) where (h,k) is vertex.

h = 1 and k + a = -1

also, we have : a = 3/2 (as calculated above)

so, k + 3/2 = -1

k = -1 - 3/2

k = -5/2

so, vertex = (1, -5/2)

thus : y-k = 4a(x-h)^2

y + 5/2 = 6(x-1)^2

hmm! doesn't seem right.

Answer 23

@ganeshie8 - any ideas?

Answer 24

If you have the equation of a parabola in vertex form y = a(x - h)2 + k, then the vertex is at (h, k) and the focus is (h, k + 1/(4a)).

therefore, focus = (1,-1) ; gives ...

h = 1 and k + 1/(4a) = -1

k + 1/6 = - 1

k = -7/6
h = 1

vertex = (1,-7/6)

So, equation : y = 3/2 ( x - 1 )^2 - 7/6

y + 7/6 = 3/2 (x-1)^2

Answer 25

naah! am confused! :(

not sure how to do this problem.

Answer 26

I will have to ask my teacher for the solution...! will post the sol. tomorrow!