Confirmation

[TrueTears]

See attached Q 2e).

Is what I did correct?

Lower half of semi-cricle has general formula (can be easily derived from by solving for y)

Subbing in coordinates of A and D yields 2 equations

[1]

[2]

The gradient is "undefined" at D, because (change in x) = 0



so when

solving for h yields

subbing in (subbing in will yield same results) into equations [1] and [2] and solving simultaneously for k and r yields and

equation is

Further info: You do not need to do part a and b or d for part e). The answer to c) is , but I don't think that is needed.

Many thanks all!

[TrueTears]

Also 1 more thing, how would you do part d). I've got a lot of working, 3 simultaneous equations, solved them and found out there is a parabola that fits into the conditions =S, but question says show its IMPOSSIBLE. Have discussed with kamil, the maths seems all good, but for reason we are stuck on how to prove it's impossible for a parabola to exist.

The parabola I got was negative parabola. The maths (I don't think) has any problems with it, but when you sketch it, there is indeed a "cusp" at x = -3, ie a disjoint part kinda like the one you see in y = |x|. But how do I show this algebraically?

Thanks once again!

[Mao]

part d,

It can be shown geometrically that any two gradients of a parabola makes an angle less than 180 degrees.

The straight stretch is verticle, and the gradient at point A is positive. That is, the two gradients do not cross below y=-126 (or they make an 'reflexive' angle). Hence, no parabola can join to smoothly to both curves.

part e,
It can be shown that it is also impossible to fit a semicircle smoothly to it with the same reasoning as above. [However, it is possible to fit an arc spanning to it]

However, if we need to fit it there, AD must be the diameter. Since they are 17 units across, radius is hence 8.5
The center is hence (5.5, -126)