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C4 - Points at which Tangent to Curve Parallel to Y-Axis

OK, I'm doing some last-minute homework for school tomorrow and am stuck on a part of a question.

The equations for the curve is: (1 + x)(2 + y) = x^2 + y^2
So it's kind of like a circle I guess.
The first part is to find dy/dx in terms of x and y, which I've done.
But I can't remember what you're meant to do to do this part:
Show also that there are two points at which the tangents to this curve are parallel to the y-axis.

The y-axis is the line x = 0, but how do you show that something is parallel to that?

Thanks in advance. :smile:
Well, it's not a circle. But ok. :wink:

If a line is parallel to the y-axis, its gradient is infinite, right? So if you have the gradient in the form dydx=a/b\frac{\text{d} y}{\text{d} x} = a/b, then b = 0 where the gradient is infinite.
Reply 2
Thanks for the quick reply.

I thought it had something to do with an infinite gradient, I just didn't know how to apply it.

By the way, what is the shape of that curve?
My graphical calculator can't seem to handle it.
alex_hk90
Thanks for the quick reply.

I thought it had something to do with an infinite gradient, I just didn't know how to apply it.

By the way, what is the shape of that curve?
My graphical calculator can't seem to handle it.

You could go and get some mystical tool on the internet to sketch it. Or look at the properties of the equation (looking for asymptotes, minima, maxima, discontinuities, etc.). You know there's at least one vertical tangent - how many are there? Is x minimum or maximum here? Are there horizontal tangents, and where, and how many? Is y maximum or minimum? Are there asymptotes? Are x and y unbounded, or is the domain/range only a subset of the set of real numbers?
Reply 4
also stuck on this question, I can get a y co-ord of (rt.15 +4)/3 but the correct solution is of rt. 13 not rt. 15 .... where might I have gone wrong?
Reply 5
dydx\frac{dy}{dx} = 2+y2x2yx1\frac{2 + y - 2x}{2y -x -1}

so I sub x=2y1x = 2y - 1 into main equation - (1+2y1)(2+y)=x2+y2(1+2y-1)(2+y)=x^2 +y^2

can anyone help please?
Here's what it looks like:
Reply 7
deosnt really help though...
Sambo2
dydx\frac{dy}{dx} = 2+y2x2yx1\frac{2 + y - 2x}{2y -x -1}

so I sub x=2y1x = 2y - 1 into main equation - (1+2y1)(2+y)=x2+y2(1+2y-1)(2+y)=x^2 +y^2

can anyone help please?

You still have an x^2 on the right - rewrite that as (2y - 1)^2. This gives you a quadratic equation in y. I get 3y^2 - 8y + 1 = 0. If you solve that (correctly), you get (4 + root 13)/3. It sounds like your slip was purely algebraic...
Reply 9
but 3y28y+1=03y^2 - 8y +1 = 0 gives (3y4)2+1=16(3y - 4)^2 + 1 = 16

so (3y4)=15(3y - 4) =\sqrt 15


So how do you get (4+15)/3(4 + \sqrt15)/3
Reply 10
Sambo2
but 3y28y+1=03y^2 - 8y +1 = 0 gives (3y4)2+1=16(3y - 4)^2 + 1 = 16

so (3y4)=15(3y - 4) =\sqrt 15


So how do you get (4+15)/3(4 + \sqrt15)/3


There the same thing? :yep:

(3y4)=15(3y - 4) =\sqrt 15 = (4+15)/3(4 + \sqrt15)/3

Add the 4 to the right
then divide by 3

If that is your question :confused:
Reply 11
sorry no I meant (4+13)/3(4 + \sqrt13)/3
Reply 12
can a maths guru help put this question to bed once and for all... please!! :biggrin:
Reply 13
@ sambo

2y = x + 1

2y(2+y) = (2y-1)^2 + (y^2)

multiply all that out, collect terms

3(y^2) -8y +1 = 0

y = 8 + or - sqrt(8^2-4(1x3))

= (8 plus or minus root 52) / 6

= (8 plus or minus 2 root 13) / 6

[since]

= (4+or- root 13) / 3

[divide top and bottom by 2]

peace