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C4 maths implicit differentiation help

' The two points on the curve x^2 + y^2 -xy = 84 where the gradient of the curve is 1/3 are A and B. Find the equations of the the tangents at A and B. '

I differentiated the equation to get what I think is correct : 2x + 2ydy/dx -xdy/dx -y = 0 .
It's just from here I seem to have no clue what to do, I make it equal to 1/3 but it doesn't seem to work out for me.

Any help is really appreciated so thank you
(edited 6 years ago)
You have to have only dy/dx on one side so it's equal to a implicit equation involving x's and y's and make that equal to 1/3. Then find y in terms of x and put that into the original equation to find the y coordinates then use y=mx+c to find c.
(edited 6 years ago)
Simplify
Dy/dx(term n + term m) = 2x-y
I have left the missing bits then rearrange and put in your values at point P to get your values for the gradient.
Reply 3
Original post by Vikingninja
You have to have only dy/dx on one side so it's equal to a implicit equation involving x's and y's and make that equal to 1/3.


yep so I tried to do what I think you mean where i got : dy/dx = (y-2x)/(2y-x) = 1/3. But from this I can't seem to get rid of the x or y for me to then be able to get a coordinate of some sort?? I just don't know what I'm not seeing
Original post by olivia099
' The two points on the curve x^2 + y^2 -xy = 84 where the gradient of the curve is 1/3 are A and B. Find the equations of the the tangents at A and B. '

I differentiated the equation to get what I think is correct : 2x + 2ydy/dx -xdy/dx -y = 0 .
It's just from here I seem to have no clue what to do, I make it equal to 1/3 but it doesn't seem to work out for me.

Any help is really appreciated so thank you
You would probably get more responses if you posted in the maths forum: http://www.thestudentroom.co.uk/forumdisplay.php?f=38

(Tagging Vikingninja in case he as abiltiy to move it...)

Edit: also worth pointing out Math forum rules: you need to post what you've done so far if you want people to respond.
Original post by olivia099
yep so I tried to do what I think you mean where i got : dy/dx = (y-2x)/(2y-x) = 1/3. But from this I can't seem to get rid of the x or y for me to then be able to get a coordinate of some sort?? I just don't know what I'm not seeing


Multiply both sides by 2y-x so then you can find x in terms of y (sorry meant to say that, you want the original equation to only contain y's).
Reply 6
Original post by DFranklin
You would probably get more responses if you posted in the maths forum: http://www.thestudentroom.co.uk/forumdisplay.php?f=38

(Tagging Vikingninja in case he as abiltiy to move it...)

Edit: also worth pointing out Math forum rules: you need to post what you've done so far if you want people to respond.


Ok apologies this is the first time i've posted anything so I had no clue
Original post by olivia099
yep so I tried to do what I think you mean where i got : dy/dx = (y-2x)/(2y-x) = 1/3. But from this I can't seem to get rid of the x or y for me to then be able to get a coordinate of some sort?? I just don't know what I'm not seeing
Don't do that - it's unnecessary complications.

From 2x + 2ydy/dx -xdy/dx -y = 0 simply replace dy/dx by 1/3. It's very easy to get y in terms of x, then sub into x^2+y^2-xy = 84 and you'll be done in a couple of lines.
Reply 8
Original post by Vikingninja
Multiply both sides by 2y-x so then you can find x in terms of y (sorry meant to say that, you want the original equation to only contain y's).


THANK YOU! EPIPHANY! I just didn't realise I could just use the value for x in terms of y but yeah I completely understand that now. Thanks for the help :laugh::h:
Original post by DFranklin
You would probably get more responses if you posted in the maths forum: http://www.thestudentroom.co.uk/forumdisplay.php?f=38

(Tagging Vikingninja in case he as abiltiy to move it...)

Edit: also worth pointing out Math forum rules: you need to post what you've done so far if you want people to respond.


Sorry I'm not the right section, tagging another guy who may be able to move it.
@RDKGames
Original post by olivia099
yep so I tried to do what I think you mean where i got : dy/dx = (y-2x)/(2y-x) = 1/3. But from this I can't seem to get rid of the x or y for me to then be able to get a coordinate of some sort?? I just don't know what I'm not seeing


(y-2x)/(2y-x) = 1/3

Using this you should be able to get y in terms of x. y = 5x
sub that back into the ORIGINAL equation (the x^2 + y^2 -xy = 84) and you should get two answers for x. then sub back in for the y cordinates
Reply 11
Original post by Formless
(y-2x)/(2y-x) = 1/3

Using this you should be able to get y in terms of x. y = 5x
sub that back into the ORIGINAL equation (the x^2 + y^2 -xy = 84) and you should get two answers for x. then sub back in for the y cordinates



Yeah that's why I eventually did and the answer i got was right - thanks :h::h:
Reply 12
Original post by Vikingninja
You have to have only dy/dx on one side


No you don't - this is an unnecessary complication; see DFranklins post.

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