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Uh huh... The focus!! Thanks... :D

First ... x^2 - 6x = 4y + 3 Then x^2 - 6x + 9 = 4y + 3 + 9 (x - 3)^2 = 4(y + 3) From here I went .... (x - 3)^2 = 4(y - (-3) ) V: (3,-3) Where have I gone wrong?

Hmmmm.... I thought I'd done this... I'll put my working out here...

Hello All. :) The question is... (i) Find the coordinates of the focus and (ii) the equation of the directrix of a) x^2 - 6x - 4y - 3 = 0 I keep getting (3,-3) for i) but the answers have (3,-2)... Which is correct?? Could someone please help me??? Thanks in advance. :)

Yeah, I see... Do you have any you could share with me??

hehe! Very true...

Hello All... Is there much difference between the current and previous Business Studies syllabus? Would appreciate others answers. Thanks. :D

Really?? Could be an idea... :D

And after walking out, you discuss the exam with others and find out they have a different answer to you! :)

Thanks heaps... Appreciate that. :)

Sorry, can someone please explain a) fully? I'm just not getting it. Thanks. :)

The Preliminary Book??

Yup, ok. thanks. :)

Thanks Man... (No worries about sqrt signs). :)

Thanks... For the very detailed explanation... :) I do, what the parts are but not sure what do with them!

Sorry can you please explain a further? :)

Yeah, ok then... Thanks. :)

Hello All. :D t's a bit of a dumb question. But appreciate any/all help. Q(2aq, aq^2) lies on the parabola x^2 = 4ay. a)Find the equation of the focal chord through Q. b)Prove that the length of the latus rectum is 4a. Thanks in advance. :)

Yup, I get yah... Thanks heaps. :)

Can someone please help me? :)