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Quadratic Polynomial Help! From Cambridge. (1 Viewer)
- Thread starter JasonNg
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Nooblet94
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- 2012
Part a)
<a href="http://www.codecogs.com/eqnedit.php?latex=y=kx^2 \textrm{ has tangent }8x-y-4=0\\ ~\\ $Solving simultaneously, we have $ 8x-kx^2-4=0\\ $Now, for the line to be a tangent, it intersects the parabola at one point, so there is only one solution to the above equation, i.e. $\Delta =0\\ ~\\ \therefore 8^2-4\times -k \times -4=0 \Rightarrow k=4" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=kx^2 \textrm{ has tangent }8x-y-4=0\\ ~\\ $Solving simultaneously, we have $ 8x-kx^2-4=0\\ $Now, for the line to be a tangent, it intersects the parabola at one point, so there is only one solution to the above equation, i.e. $\Delta =0\\ ~\\ \therefore 8^2-4\times -k \times -4=0 \Rightarrow k=4" title="y=kx^2 \textrm{ has tangent }8x-y-4=0\\ ~\\ $Solving simultaneously, we have $ 8x-kx^2-4=0\\ $Now, for the line to be a tangent, it intersects the parabola at one point, so there is only one solution to the above equation, i.e. $\Delta =0\\ ~\\ \therefore 8^2-4\times -k \times -4=0 \Rightarrow k=4" /></a>
b) is the same method as above, except the equation of the parabola isn't given. You can easily deduce that the equation is of the form y=kx^2 since it says that the vertex is at the origin and it has a vertical axis.
<a href="http://www.codecogs.com/eqnedit.php?latex=y=kx^2 \textrm{ has tangent }8x-y-4=0\\ ~\\ $Solving simultaneously, we have $ 8x-kx^2-4=0\\ $Now, for the line to be a tangent, it intersects the parabola at one point, so there is only one solution to the above equation, i.e. $\Delta =0\\ ~\\ \therefore 8^2-4\times -k \times -4=0 \Rightarrow k=4" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=kx^2 \textrm{ has tangent }8x-y-4=0\\ ~\\ $Solving simultaneously, we have $ 8x-kx^2-4=0\\ $Now, for the line to be a tangent, it intersects the parabola at one point, so there is only one solution to the above equation, i.e. $\Delta =0\\ ~\\ \therefore 8^2-4\times -k \times -4=0 \Rightarrow k=4" title="y=kx^2 \textrm{ has tangent }8x-y-4=0\\ ~\\ $Solving simultaneously, we have $ 8x-kx^2-4=0\\ $Now, for the line to be a tangent, it intersects the parabola at one point, so there is only one solution to the above equation, i.e. $\Delta =0\\ ~\\ \therefore 8^2-4\times -k \times -4=0 \Rightarrow k=4" /></a>
b) is the same method as above, except the equation of the parabola isn't given. You can easily deduce that the equation is of the form y=kx^2 since it says that the vertex is at the origin and it has a vertical axis.