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Annoyingly hard q (1 Viewer)
- Thread starter darlking
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When solving for the points of intersection
and
^2=r^2)
We have
^2=r^2\\\\\Rightarrow y^2+(1-2c)y+c^2-r^2=0)
Since the circle is tangent symmetrically the two points of contact must have the same y-value. Hence there is only one solution in y to the equation above. If there is only one solution the discriminant must be zero.
^2-4(c^2-r^2)=0\\\\\Rightarrow 4c=1+4r^2)
When you actually solve the equation for y using the quadratic formula and the result in (i) you get
However, from the diagram it is pretty obvious thay y must be positive so
and
We have
Since the circle is tangent symmetrically the two points of contact must have the same y-value. Hence there is only one solution in y to the equation above. If there is only one solution the discriminant must be zero.
When you actually solve the equation for y using the quadratic formula and the result in (i) you get
However, from the diagram it is pretty obvious thay y must be positive so
Makematics
Well-Known Member
you couldnt find any solutions to a 2012 hsc paper? :S check out itute or something next time lol
luvglee4lyfe
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Hey Makematics, what's itute?
SpiralFlex
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They have 'sample' VCE mainly but also HSC Mathematics solutions. Here: http://www.itute.com/download-free-vce-maths-resources/free-maths-exams/
I have not checked it out yet so don't know how trustworthy they are.
luvglee4lyfe
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Thank you!