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Maths equations help (tangents)

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acroviak | 22:06 Mon 29th Mar 2010 | Quizzes & Puzzles
4 Answers
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!
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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...
22:19 Mon 29th Mar 2010
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
Question Author
Awesome thanks for that one. Any idea on the second one i cant solve?
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
Question Author
Thanks again!

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

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