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

Step 2. Equate the coordinates at the point of contact :

Step 3. Subst eqn1 into eqn2 to eliminate

(ii) From step 1 above:

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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