# Q16 (1 Viewer)

#### uart

##### Member
A bit late but I've only just got to look at this exam today. Using discriminates was a good solution, but another way to tackle 16c is to equate both the gradients and the coordinates at the point of contact. One advantage of this approach is that it makes the second part basically trivial.

Use calculus and y' = 2x for the parabola gradient, but just use m2 = -1/m1 (where m1 is the gradient of the radius from 0,c to x,y) for the gradient of the circle.

Step 1. Equate the gradients at the point of contact : 2x = x/(c-y), so c-y = 1/2

Step 2. Equate the coordinates at the point of contact : y + (c-y)^2 = r^2

Step 3. Subst eqn1 into eqn2 to eliminate y. This gets you the solution in one line.

(ii) From step 1 above: c = y + 1/2. Point of contact must have y>0 which immediately gives the desired inequality.

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#### RealiseNothing

##### what is that?It is Cowpea
y = c - 1/2
and knowing that the equation of the parabola is y = x^2 which means y is always positive (concave up parabola), we deduce that y.. must be positive

therefore,

c - 1/2 > 0
and then c > 1/2

Very easy, this took me about 5 minutes in class (I'm in Yr 11 atm) but under pressure I could DEFINITELY understand how people would shit bricks
You would have to justify why it is a strict inequality, since $\bg_white y\geq 0$

#### drlisir

##### New Member
16(c)I.
Solving simultaneously: x^2 +(y-c)^2=r^2 and y=x^2 for the coordinates of the touching points,
Hence : y+(y-c)^2=r^2, which is a quadratic equation in y.
Buy the touching points are equal in y coordinates, so discriminant of the equation equals to 0.
So (1-2c)^2 - 4 (c^2 - r^2 = 0. Giving 4c = 1 + 4 r^2.

(II). From figure, c > r (OC is opposite to the right angle in the triangle)
Using result in (I), 4c = 1 + 4 r^2 > 4r
So 4 r^2 - 4 r + 1 > 0
Or (2 r - 1)^2 > 0
2 r - 1 > 0
r > 1/2
But c > r, so c > r > 1/2, c > 1/2.

#### uart

##### Member
You would have to justify why it is a strict inequality, since $\bg_white y\geq 0$
The question states that the circle touches at two points located symmetrically, so it really can't touch at y=0 (or it would at least touch at an odd number of points). Of course it wouldn't hurt to state an explicit justification, but in a 2 unit paper I highly doubt that it would be necessary to get the mark.

It's interesting to look at the limiting case where c=1/2. You can show that not only do the coordinates and the first derivative match up, but so too do the second and third derivatives. So though the circle only touches the parabola at the one point (the origin), it is quite a snug fit in there.

#### drlisir

##### New Member
as c>0, so 1+4 r^2>0
Or 4 r^2 > -1
How can you deduce that r> 1/2?

#### uart

##### Member
as c>0, so 1+4 r^2>0
Or 4 r^2 > -1
How can you deduce that r> 1/2?
Do you mean given that c > 1/2 show that r > 1/2?

Rearrange 4c = 1 + 4 r^2 to,

$\bg_white r = \sqrt{c-0.25}$.

And note that when c=0.5 that r=0.5,

and that $\bg_white \sqrt{\cdot}}$ is an increasing function.

Edit: Re-reading your question I'm not sure I really answered what you asked. Anyway, it's easiest to first prove the c = y + 1/2, and therefore that c>1/2, and then if you wish prove that this implies r>1/2.

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#### G0

##### New Member
whats the cut off for band 6