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# Curves Defined Implicitly Help!!!!! watch

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1. Hi,

Im stuck on these 2 questions on the OCR board P3:

1) x^2 - 2xy + 2y^2 = 5 find the tangent to the curve at the point (1,2)

2) find the points on the curve 4x^2 + 2xy - 3y^2 = 39 at which the tangent is parallel to one of the axes.

Cheers

Streety
2. 1) 2x - 2[y+x(dy/dx)] + 4y(dy/dx) = 0
Now substitute x=1 and y=2 to find dy/dx, i.e. the gradient of the tangent.

2) 8x + 2[y+x(dy/dx)] - 6y(dy/dx) = 0
The tangent is parallel to an axis when it touches the maximum point. So, substitute dy/dx=0 to get the equation.
5. (Original post by streetyfatb)
Hi,

Im stuck on these 2 questions on the OCR board P3:

1) x^2 - 2xy + 2y^2 = 5 find the tangent to the curve at the point (1,2)

2) find the points on the curve 4x^2 + 2xy - 3y^2 = 39 at which the tangent is parallel to one of the axes.

Cheers

Streety
This is probably wrong, but no one else's answered and I'm bored.

1)
x^2 - 2xy + 2y^2 = 5 differentiate
2x-2y -2x*(dy/dx)+4y(dy/dx)=0 substitute (1,2)
2-4-2(dy/dx)+8(dy/dx)=0
6dy/dx=2
dy/dx=1/3
Tangent: y=mx+c so m=1/3
y=x/3+c substitute (1,2)
2=1/3+c
c=5/3
y=x/3+5/3
3y-x=5 is the tangent.

2)

4x^2 + 2xy - 3y^2 = 39
8x+2y+2x(dy/dx)-6y(dy/dx)=0
dy/dx(2x-6y)=-2y-8x
dy/dx=(8x+2y)(6y-2x)
Tangent parallel to axis when gradient is zero or infinite. It is therefore parallel to axis when: y=-4x (i.e. dy/dx is 0, parallel to x axis), when x,y=0 (gradient infinite, parallel to y axis) and when y=3x, again gradient infininte.
6. Cheers guy, still not with qu 2 yet so if u could explain 2 me it would be much appricated.

Cheers

Streety
7. The gradient is defined as tan(m), where m is the angle the tangent makes with the x-axis.

For the line to be parallel to the x-axis, it must touch the maximum point of the curve, i.e. when m=180. That is: dy/dx = tan(m) = tan180 = 0.

For the line to be parallel to the y-axis, it must make a right angle with the x-axis. That is: dy/dx = tan(m) = tan90 = undefined/infinity.

I don't think you're expected to use the second case though, since the question asked for an equation of the tangent parallel to one of the axes.
8. (Original post by streetyfatb)
Cheers guy, still not with qu 2 yet so if u could explain 2 me it would be much appricated.

Cheers

Streety
You start off with: 4x^2 + 2xy - 3y^2 = 39
Then differentiate everything with respect to x: 8x+2y+2x(dy/dx)-6y(dy/dx)=0
You know factorise the dy/dx parts: 8x+2y+(2x-6y)=0
Then shift the other terms to the other side: (2x-6y)(dy/dx)=-2y-8x
Then divide through by (2x-6y) to get dy/dx by itself: dy/dx=(-8x-2y)/(2x-6y)
Then get rid of the minus signs: dy/dx=(8x+2y)/(6y-2x)
You've now obtained an expression for dy/dx.
For dy/dx to be parallel to the axis, there are two cases: parallel to the x axis, and parallel to the y axis. Parallel to the x axis means that no matter what the shift in x is, the shift in y is zero. Hence dy/dx=0. For this to happen, (8x+2y)/(6y-2x) has to be zero. i.e. y=-4x.
If the tangent is parallel to the y axis, that means for no x increase at all, you get as large a y increase as you like. This means dy/dx is infinite. So (8x+2y)/(6y-2x) is inifite. That means 6y-2x has to be zero; that can happen in two ways: either the'yre both zero, or x=3y.
So in total, you've got three set of solution: one infinite set of solutions where y=-4x, one infininte set where x=3y, and one isolated case, where x=y=0
9. And a bit more... The expression for the gradient is zero when y=-4x and infinite when x=3y but it's important not to forget that not all points that satisfy these also satisfy the original equation.

4x^2 + 2xy - 3y^2 = 39

Straight away x=y=0 does not satisfy this so this cannot be one of the points.

Subbing in x=3y for infinite gradient:
4(3y)^2 + 2(3y)y - 3y^2 = 39
36y^2 + 6y^2 - 3y^2 = 39
39y^2 = 39
y = +-1 => x = +-3

4x^2 + 2x(-4x) - 3(-4x)^2 = 39
4x^2 - 8x^2 - 48x^2 = 39
-52x^2 = 39
This will lead to an imaginary solution so dy/dx =/= 0 so the tangent is never parallel to the x axis.

So the points are (3,1) and (-3,-1) where we have infinite gradient and the tangent is parallel to the y axis.

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