My AS further maths class and I have finished the scheme for learning for the year, leaving oodles of time for review, recap and plenty of practice. Currently we are revisiting topics that they either didn’t score too well on the last mock in or that they have requested we look at again due to lacking confidence.

Today we were looking at conic sections, by request of the students, and this past paper question caught my eye:

We had recapped the topic quickly together then the students attempted so past paper questions. This one was one they rook to well, all competing the first 2 sections quickly and without trouble. Likewise, part c was mostly uneventful – bar one or two silly substitution errors and someone missing a difference of two squares factorisation.

Then they got to part d.

This caused them a little bit of worry so we worked together on it as a class, after we’d made sure everyone had a, b and c right.

Here’s what the board looked like when we’d done:

I asked the class what we should do to start, one suggested drawing a diagram all this nagging about always drawing a diagram, especially if your stuck is paying off! We drew it, but it didn’t help much:

Then one said, “if they’re parallel then one gradient is -1 over the other” – I refrained from scolding and calmly said “indeed, those gradients will be negative reciprocals of each other”. I think asked what the gradients were and quickly had “y1 – y2 over x1 minus x2” thrown at me.

So we worked out the gradient of the line join in n to the origin:

Then the gradient of PQ:

One then suggested we put the equal to each other, but he was corrected by another student before I could react. So we set one equal to the negative reciprocal of the other and solved:

A lovely question, with a lovely neat answer and a load of fun algebra on the way. I enjoyed watching the students tackle it and was glad I didn’t need to put too much input in myself.

We then discussed the answer and the class expressed surprise that 1 was the final answer and that a complicated algebraic journey could end so simply. It was a nice discussion and a nice way to start the day, and the week.

*This post was cross posted to Cavmaths and One Good Thing.*

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Nice one. Considerations of symmetry will show that one of the values of p^2.q^2 is 1. This is a start at least! Then looking at a diagram and thinking about moving the point Q one can see that there will only be one place which gives the required orthogonality…………

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