Submitted by mahmed103 on Sun, 04/28/2019 - 10:48
Forums:
For part (ii) :
the solution states that the different cases that arise from the shortest distance to a circle and a parabola are:
a) p ≤ 2a
b < p < 2a ⇒ shortest distance = p - b
p ≤ b < 2a ⇒ shortest distance = 0 (as circle must pass through the origin and so does the parabola, and the circle intersects the parabola when p < b )
b ≤ p = 2a ⇒ shortest distance = 2a - b
>>>>should the last condition actually be b < p = 2a, since if b = p then we have that the parabola and the curve passes through the origin which would give shortest distance = 0?
Any help would be much appreciated!
Considering cases
Also, why do you consider p ≤ 2a and p > 2a separately for the circle and parabola?
Just checking
Just to make sure - this isn't a question that we have done on this site? (Just so that I know I don't have to find it and update the solutions!).
Which solution have you been looking at?
Second part of (ii)
I am going to assume that you have shown the first part of (ii), that is:
\[
\text {if } \frac p a \lt 2 \text{ then shortest distance is } p\\
\text {if } \frac p a \ge 2 \text{ then shortest distance is } 2\sqrt{a(p-a)}
\]
For the circle then the shortest distance will be $0$ is the circle and parabola intersect and otherwise will be $b$ less than the shortest distance between the centre of the circle and the parabola.
First case: let $\frac p a \lt 2$. Then we have:
\[
\text{if } p \le b \text{ the circle and parabola intersect so shortest distance is } 0\\
\text{if } p>b \text{ the shortest distance is } p-b\\
\]
Then if $\frac p a \ge 2$ we have:
\[
\text{if } 2\sqrt{a(p-a)} \le b \text{ the circle and parabola intersect so shortest distance is } 0\\
\text{if } 2\sqrt{a(p-a)}>b \text{ the shortest distance is } 2\sqrt{a(p-a)}-b\\
\]
Mark scheme
You might find the 2014 STEP 3 mark scheme useful.
Mistake
Hi, sorry for not replying earlier (I was in school)
The solutions provided on integral stated that a possible case gave 2a - b. The mark scheme and solution you provided confirmed that this is most likely a mistake.
Thanks!