Maths equations help (tangents)

[acroviak]

Hey.

Couple of questions im stuck with.

I've found my cartesian equation from two parametric equations, and they are correct. But then theres some tangent questions i need some help with! :D

The equation is: 8 x = y ^ 2

Questions.
- The coordinates of the points where y = x - 6 cuts the curve

- The value of K for which y = 2 x + K is a tangent to the curve.



{Incase you need the initial parametric equations, they are: x = 2 t ^ 2, y = 4 t}


Thanks alot!

[Buenchico]

Intersection:
At the point where y = x -6 cuts the curve BOTH the equations must be true
i,e. y = x - 6 AND 8x = y^2

So replace y with (x - 6) in the equation for the curve. That gives
8x = (x - 6)^2

Expand the right hand side to give
8x = x^2 - 12x +36

That's a quadratic equation, which is easiest to solve if one side equals zero. So subtract 8x from both sides (and swap left to right) to give
x^2 -20x + 36 = 0

Factorise to get
(x - 18)(x - 2) =0
<=> x = 18 OR x =2

That's provided the x coordinates for the points where the intersections occur. Substitute into either equation (since both are true) to find y.
If x = 18, y = x - 6 <=> y =12
So one intersection occurs at (18,12)
If x = 2, y = x -6 <=> y = -4
So the other intersection occurs at (2,-4)

Chris

[acroviak]

Awesome thanks for that one. Any idea on the second one i cant solve?

[Buenchico]

Thanks for the reply. I'm very tired and I've got to get up early, so I've not really got time to think about the second part, but here's a starting point for you to investigate:

The gradient of the tangent y = 2x + K = 2

So that means that you're seeking a point on the curve where the gradient is also 2. Hence dy/dx must be equal to 2.

Actually, looking at it again, it might be better to find dx/dy (which must be equal to the reciprocal of 2, i.e. ½)

x =1/8 y^2
gives dx/dy = ¼y
<=> ¼y = ½
<=> y = 2

Since y = 2 and, on the curve, 8x = y^2
<=> x = ½

Substituting into y = 2x + K we get
2 = 1 + K
<=> K = 1

Oh, I think I might have done it anyway - but that working is UNCHECKED. It might turn out to be RUBBISH! I'll leave it to you to check it!

Chris

[acroviak]

Thanks again!

Just checking the second one now, but it looks about right.