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STEP Support Programme

S3 Q3 2014 (Co-ordinate geom)

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!

Also, why do you consider p ≤ 2a and p > 2a separately for the circle and parabola?

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?

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:

\[
\text{if } 2\sqrt{a(p-a)} \le b \text{ the circle and parabola intersect so shortest distance is } 0\\
\text{if } 2\sqrt{a(p-a)}>b \text{ the shortest distance is } 2\sqrt{a(p-a)}-b\\
\]

You might find the 2014 STEP 3 mark scheme useful.

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!