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?
the other end of the locus mmbblbbmmb .... yeah
this is a parabola; opens to the left; centered at origin right
(2,8)
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.
ooohhhh; so its like a degenerate latus rectum :)
well you can say latus rectum is one kind of a focal chord
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
err... (8,0)
they say rise of run; which is y/x...but then they stick x in front to confuse the issue lol
lets create a line from (2,-8 to 8,0 and see where these meet)
2,-8 -8,0 ---- -6,8 ; slope = -4/3
slope = 4/3....
0 = (4/3)(8) + b y = (4/3)x -32/3 y = sqrt(32x)
32x = 16x^2 -256x +1024 0 = 16x^2 -224x + 1024; right?
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)
yes; 32,32 :)
apparently i had the line right; but mighta got messed up in solving the system i created :)
y^2 = 32x 3y = 4x - 32 (-8) y^2 -24y = 256 y^2 - 24y - 256 = 0 ; thats the one lol
y = -8, y = 32 3(32) + 32 ----------- = x 4 3(8) + 8 = x 24 + 8 = x = 32
your way loks faster tho...
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