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Differentiation problem watch

1. (1+2)(2+y) = x^2 + y^2

find dy/dx in terms of x and y (Done that bit)

dy/dx = 2x - 2 - y / x + 1 -2y

then find the gradient of the above equation at each of the two points where the curve meets the y-axis (Also done this bit)

dy/dx = 4/3 and -1/3

then show also that there are two points at which the tangents to this curve are parallel to the y-axis. Not got much of a clue on this bit could someone help out? A gradient parallel to the x axis is 0..but then whats the one for parallel to y? because then i could substitute that value in as dy/dx to calculate a value in terms of x's and y's surely? and they would be the points..?
2. When the tangent is parallel to the y-axis, dy/dx = infinity.
3. Parallel to x-axis => dy/dx = 0
Parallel to y-axis => dx/dy = 0
4. Thankyou I've got the value for x using that method

5 +- 2Root13 / 3

but now struggling for Y. To get x i did dy/dx = 0. therefore x+1-2y = 0
and x+1 / 2 = y which i then substitued into the first equation. to get a quadratic.

Now do i just do 2y - 1 = x? because that doesnt seem to work out for me when i substitute it in..please help again Thanks

Nevermind, simple maths error..duh :\ and only 5 or so weeks to the exam

Thanks for ya help

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