A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
If (2,8) is the coordinates of an end point of a focal chord of y^2=4ax, then the coordinates of the other end are?
anonymous
 5 years ago
If (2,8) is the coordinates of an end point of a focal chord of y^2=4ax, then the coordinates of the other end are?

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the other end of the locus mmbblbbmmb .... yeah

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this is a parabola; opens to the left; centered at origin right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no 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.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ooohhhh; so its like a degenerate latus rectum :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well you can say latus rectum is one kind of a focal chord

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0We 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0they say rise of run; which is y/x...but then they stick x in front to confuse the issue lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets create a line from (2,8 to 8,0 and see where these meet)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02,8 8,0  6,8 ; slope = 4/3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00 = (4/3)(8) + b y = (4/3)x 32/3 y = sqrt(32x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.032x = 16x^2 256x +1024 0 = 16x^2 224x + 1024; right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0http://www3.wolframalpha.com/input/?i=y^2+%3D+32x+and+y+%3D+4x%2F3++32%2F3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0apparently i had the line right; but mighta got messed up in solving the system i created :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y^2 = 32x 3y = 4x  32 (8) y^2 24y = 256 y^2  24y  256 = 0 ; thats the one lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = 8, y = 32 3(32) + 32  = x 4 3(8) + 8 = x 24 + 8 = x = 32

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your way loks faster tho...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.