Question

If (2,-8) is the coordinates of an end point of a focal chord of y^2=4ax, then the co-ordinates of the other end are?

Answer 1

the other end of the locus mmbblbbmmb .... yeah

Answer 2

this is a parabola; opens to the left; centered at origin right

Answer 3

(2,8)

Answer 4

no you are confusing it with latus rectum. thats not what it is... its a line joinin the point (2,-8) to the other end of the parabola, PASSING through the focus.

Answer 5

ooohhhh; so its like a degenerate latus rectum :)

Answer 6

well you can say latus rectum is one kind of a focal chord

Answer 7

We would have to plug in the coordinates to calibrate the parabola maybe?

-8^2 = 4a(2)

64 = 8a

a = 8; the focal point is at 0,8

Answer 8

err... (8,0)

Answer 9

they say rise of run; which is y/x...but then they stick x in front to confuse the issue lol

Answer 10

lets create a line from (2,-8 to 8,0 and see where these meet)

Answer 11

2,-8
-8,0
----
-6,8 ; slope = -4/3

Answer 12

slope = 4/3....

Answer 13

0 = (4/3)(8) + b

y = (4/3)x -32/3
y = sqrt(32x)

Answer 14

32x = 16x^2 -256x +1024

0 = 16x^2 -224x + 1024; right?

Answer 15

the parametric point of a parabola is (at^2,2at)
to suit the point (2,-8) we have, (8.(-1/2)^2, 2.8.(-1/2))
thus t=-1/2 lets call this t1, and the point at the other end of the parabola is \[(at _{2} ^{2},2at _{2}^{})\]
t1*t2=-1 so t2=2
so the other point is (8.2^2,2.8.2)
which is (32,32)

Answer 16

http://www3.wolframalpha.com/input/?i=y^2+%3D+32x+and+y+%3D+4x%2F3+-+32%2F3

Answer 17

yes; 32,32 :)

Answer 18

apparently i had the line right; but mighta got messed up in solving the system i created :)

Answer 19

y^2 = 32x

3y = 4x - 32 (-8)

y^2 -24y = 256

y^2 - 24y - 256 = 0 ; thats the one lol

Answer 20

y = -8, y = 32

3(32) + 32
----------- = x
4

3(8) + 8 = x

24 + 8 = x = 32

Answer 21

your way loks faster tho...