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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: josh92012 on June 01, 2016, 09:48:15 am
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The line with equation kx + y = 3 is a tangent to the curve with equation y= kx^2 + Kx - 1 . Find the value of k
What's the answer and how do u work it out
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The tangent to a curve at a point is the line which intersects the curve AND has the same gradient as the curve at that point. So the first thing to do is to work out where this line and curve meet:
kx + y =3 ____<1>
y = kx^2 + kx - 1_____<2>
<2> into <1>: kx + kx^2 + kx - 1 =3
k(2x+x^2)=4____<3>
The next step is to understand that the gradient of the line is the gradient of the curve at the point at which they meet.
the gradient of the line is -k, as seen by expressing <1> as y = -kx + 3. The gradient of the curve is given by:
dy/dx = 2kx+k.
so now the question is when does 2kx+k equal -k?
2kx +k = -k
x = (-k -k)/(2k)
x = -1
substituting this back into <3> gives:
k(2*(-1)+(-1)^2)=4
k*(-2+1)=4
k*-1=4
k=-4