# Further Maths Rectangular Parabolas

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#1
Hi everyone,

I had this question on a recent mock I did, and I was wondering about the best way to do it. It's only part b (part a is simple enough); apologies for it being sideways.

I did manage a solution in the end by doing a very convoluted and irritating method of setting the normal and the parabola equal to each other, and going through some obnoxious algebra to get to the coordinates of Q. However, this took me far too long in the exam and ate away at my time. Is there a better/quicker way of doing this question? I tried to use the equation of the normal we were given (as that is generally what the examiners want us to do), but I couldn't get anywhere.
Last edited by Clarkest; 1 year ago
0
1 year ago
#2
(Original post by Clarkest)
Hi everyone,

I had this question on a recent mock I did, and I was wondering about the best way to do it. It's only part b (part a is simple enough); apologies for it being sideways.

I did manage a solution in the end by doing a very convoluted and irritating method of setting the normal and the parabola equal to each other, and going through some obnoxious algebra to get to the coordinates of Q. However, this took me far too long in the exam and ate away at my time. Is there a better/quicker way of doing this question? I tried to use the equation of the normal we were given (as that is generally what the examiners want us to do), but I couldn't get anywhere.
Probably sub
y = c^2/x
into the equation of the normal to get a quadratic in x. The two roots should give the two points on the hyperbola but you only need midpoint, so complete the square and get
x = -b/2a
which I think is
-c(1-p^4)/2p^3
Last edited by mqb2766; 1 year ago
0
#3
(Original post by mqb2766)
Probably sub
y = c^2/x
into the equation of the normal to get a quadratic in x. The two roots should give the two points on the hyperbola but you only need midpoint, so complete the square and get
x = -b/2a
which I think is
-c(1-p^4)/2p^3
Hm, I might need a bit more guidance (your x coordinate is absolutely right). How did you get x = -b/2a?

I got a quadratic in x with coefficients A = p^3, B = c(1-p^4), and C = -pc^2, but completing the square on this is no easy task... Also, how does completing the square allow you to find the midpoint, as opposed to finding the x coordinate of Q?
0
1 year ago
#4
(Original post by Clarkest)
Hi everyone,

I had this question on a recent mock I did, and I was wondering about the best way to do it. It's only part b (part a is simple enough); apologies for it being sideways.

I did manage a solution in the end by doing a very convoluted and irritating method of setting the normal and the parabola equal to each other, and going through some obnoxious algebra to get to the coordinates of Q. However, this took me far too long in the exam and ate away at my time. Is there a better/quicker way of doing this question? I tried to use the equation of the normal we were given (as that is generally what the examiners want us to do), but I couldn't get anywhere.
Similar to what mqb said, note that if you have a quadratic x^2 + Ax + B, then the sum of the roots is -A, so the "midpoint in x" is going to be -A/2.

It's generally worth bearing in mind these relationships between roots and coefficients, as examiners often set questions where using them can save you a fair amount of time.

It's also worth noting that if you have a tangent or normal to the point (X, Y) on a curve, then if you substitute the equation of the line into the curve to get an equation f(x) = 0 whose roots are the points where the line meets the curve (i.e. what you've done), then you already know that x = X is one root of f. This often saves a fair bit of time/calculation as well.

Edit: also, for the y-coordinate here, notice that if you "swap" x and y in xy = c, nothing actually changes in the equation. On the other hand, the point (cp, c/p) becomes the point (c/p, cp); that is, p goes to 1/p. So you'd expect the y-coordinate to come out as -c(1-1/p^4)/2(1/p^3) = -c(p^3 - 1/p)/2 = -c(p^4 - 1)/2p. (Might not be right - not actually done any calculations).
Last edited by DFranklin; 1 year ago
2
1 year ago
#5
(Original post by Clarkest)
Hm, I might need a bit more guidance (your x coordinate is absolutely right). How did you get x = -b/2a?

I got a quadratic in x with coefficients A = p^3, B = c(1-p^4), and C = -pc^2, but completing the square on this is no easy task... Also, how does completing the square allow you to find the midpoint, as opposed to finding the x coordinate of Q?
You'll probably kick yourself a bit. There are a few similar ways of looking at it, but if you have a quadratic
ax^2 + bx + c
The roots are
x = -b/2a +/- *

Completing the square
a(x^2 + xb/a + c/a)
a(x+b/2a)^2 + *
where * is another constant. Its minimum (point of symmetry) is -b/2a. Its the midpoint of the roots.

Its "basic" quadratic stuff but easy to overlook.
Last edited by mqb2766; 1 year ago
2
#6
Ah, I've got it. An ingenious couple of methods - I'm not used to synoptic questions like that where they bring in a topic from a different area of the spec (i.e. roots of equations), so I never considered it. I've also never even used the fact that the minimum of a quadratic is the midpoint of its roots before... very clever. Thanks a lot guys 1
1 year ago
#7
(Original post by Clarkest)
Ah, I've got it. An ingenious couple of methods - I'm not used to synoptic questions like that where they bring in a topic from a different area of the spec (i.e. roots of equations), so I never considered it. I've also never even used the fact that the minimum of a quadratic is the midpoint of its roots before... very clever. Thanks a lot guys FWIW, I remember this topic (parabolas, hyperbolae, ellipses, etc.) as being *really* heavy on spotting tricks/relationships.

Quite a lot of it is practice - often the same (or similar) equations keep coming up, and you get used to what kinds of tricks work. Even so, it's always a fairly stern test of your algebra skills.
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