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# Parabola question (FP1 Coordinate Systems) watch

1. Hello! I have been wading through even questions in Review Exercise 1 of the FP1 book for the last 4/5 hours. I'm in a jiffy over question 62 part b).

The first part asks me to show that an equation of the normal to the parabola at is .
I found the gradient of the parabola at that point and then found its negative reciprocal to find the normal's gradient. Since I know the gradient of the line and that it passes through P, which is given, I find the formula of the normal.

Part b) states "The normal to the parabola at P meets the parabola again at Q. Find, in terms of t, the coordinates of Q."
I would have thought that you simply equate the formulas of the parabola
and the normal after changing the parabola's equation into y=... format and then just rearrange to get x on its own but it's just not working out in my calculations. I tried doing the opposite and looking to get y on its own but I didn't get the right answer.

The answer is meant to be ( a((t^2 + 2)/2)^2 , -2a((t^2 + 2)/2) ) <-- Sorry about that but LaTeX went loco

Any ideas on what I'm doing wrong? My head is steaming from work and I'm paranoid about missing something very simple. I would try to post my workings but it's an absolute mess of brackets and squares in places.
2. (Original post by reluE)
Hello! I have been wading through even questions in Review Exercise 1 of the FP1 book for the last 4/5 hours. I'm in a jiffy over question 62 part b).

The first part asks me to show that an equation of the normal to the parabola at is .
I found the gradient of the parabola at that point and then found its negative reciprocal to find the normal's gradient. Since I know the gradient of the line and that it passes through P, which is given, I find the formula of the normal.

Part b) states "The normal to the parabola at P meets the parabola again at Q. Find, in terms of t, the coordinates of Q."
I would have thought that you simply equate the formulas of the parabola
and the normal after changing the parabola's equation into y=... format and then just rearrange to get x on its own but it's just not working out in my calculations. I tried doing the opposite and looking to get y on its own but I didn't get the right answer.

The answer is meant to be ( a((t^2 + 2)/2)^2 , -2a((t^2 + 2)/2) ) <-- Sorry about that but LaTeX went loco

Any ideas on what I'm doing wrong? My head is steaming from work and I'm paranoid about missing something very simple. I would try to post my workings but it's an absolute mess of brackets and squares in places.
You've picked the harder way round, I think, in rearranging the parabola's equation; if you do that, you have to worry about the plus-or-minus right from the beginning. Since the line is really easy to get into y=(something), because you can just subtract xt from both sides, it's much simpler to substitute into the parabola's formula. You end up with a quadratic in x (with some t's floating around, but no y's because you've substituted them away) which you can solve. Remember that you already know one of the solutions (at^2) so that gives you one factor straight away, and it's easy to factorise a product of two factors if you know one of the factors

3. Not, we know at this point that must be a solution, so you can factorise for the other y coordinate. Then of course substitute it back in for the x coordinate.
4. (Original post by reluE)
[FONT=verdana]...
What'd I'd do is take the equation of the normal.

Then let q be the parameter for the points where the normal intersects the parabola.

We know the x,y co-ordinates of a point on the parabola in terms of the parameter q.

So substitute those into the equation of the normal.

You will get two possible values of q. One corresponding to your original point (q=t - knowing this can be useful in solving the equation), and one to the new point.
5. Like buses, none for ages, and then three come at once!
6. Ok thanks everyone, the problem was in expressing my equation in a form where factorising becomes possible.

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