Parabola Problem
A parabola passes through the following points
( − 1 , 1 ) , ( 0 , 0 ) , ( 2 1 , 4 1 ) , ( 1 , 1 ) , ( − 5 7 , − 5 3 ) .
It also passes through the point: ( 2 4 0 2 8 9 , b a ) , where a , b are positive coprime integers. Find a + b
The answer is 1759.
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2 solutions
How did you find the implicit equation?
I initially thought it was a standard parabola, and was confused.
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Calvin, I had to put on my lawyer hat and think about the wording of this problem very carefully, and avoid any use of the word "function". When I was young and foolish, I used "function" and "implicit equation" interchangeably. But that upsets a lot of people, and now in my old age, I've finally learned to respect the distinction.
Finding that implicit equation is a lot harder than actually solving this problem. But of course, it starts out by solving a simultaneous system of linear equations, by plugging in different points into this equation
a y 2 + b x y + c x 2 + d y + e x + f = 0
which anybody can do now to solve this problem.
Then it's a matter of fiddling around with the points for a "nice" problem. This is where a little creativity helps.
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LOL
For completeness, to find the (unique) equation of a conic section given 5 points ( p i , q i ) , we can use the equation:
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ p 1 2 p 2 2 p 3 2 p 4 2 p 5 2 x 2 p 1 q 1 p 2 q 2 p 3 q 3 p 4 q 4 p 5 q 5 x y q 1 2 q 2 2 q 3 2 q 4 2 q 5 2 y 2 p 1 p 2 p 3 p 4 p 5 x q 1 q 2 q 3 q 4 q 5 y 1 1 1 1 1 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0
This avoids having to tediously solve the linear equations (but that's what matrices are for!)
Why does this work?
Hint: What happens if the first row of the matrix is replaced by the coefficients of our system, namely a b c d e f ?
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@Calvin Lin – You got it!
@Calvin Lin – In this case you can narrow things down pretty quickly. We can write x 2 + b x y + ( b 2 / 4 ) y 2 + c x + d y + e = 0 , and quickly we see e = 0 and also c = − b (from plugging in ( − 1 , 1 ) and ( 1 , 1 ) and subtracting equations). So now we're down to two unknowns. From there you have to do a little linear algebra.
Interesting fact: there are exactly two parabolas passing through the first four points--I bet you can guess the other one! I'm guessing that, to make the problem, you started with those four points and computed the other parabola?
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@Patrick Corn – I diddled around a litte more than that, because, remember, if you just pick four points at random, it's not easy to surmise what parabolas could go through it---and the whole idea is to make the problem non-obvious, and have at least another point nearby with "simple" rational coordinates, and there's the matter of making sure that the one point with the unique y is simple enough too.
@Calvin Lin – "Why does this work?" does have the easy (but incomplete) answer that whenever 2 rows of a matrix are the same, then the determinant is 0. The equation expressed by this determinant would be 0 when (x,y) is the same as any one of the points.
However, I haven't yet made any sense of "what happens if the first row of the matrix is replaced by a, b, c, d, e, f" I'm still scratching my head with that one.
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@Michael Mendrin – Ah, the wonders of linear algebra. Let me move this discussion into a note , and see what others have to say.
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@Calvin Lin – I see your post about this interesting subject, yes, let's see what others have to say.
I still don't understand how did you got the equation of parabala. Could you explain more?
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Let
a
y
2
+
b
x
y
+
c
x
2
+
d
y
+
e
x
+
f
=
0
be the
2
n
d
order implicit equation for any conic section (of which a parabola is one kind, but that's already been taken care of for you). Then you simply plug in each of the
5
points as given into this equation, and end up with the following
a
−
b
+
c
+
d
−
e
+
f
=
0
f
=
0
1
6
1
a
+
8
1
b
+
4
1
c
+
4
1
d
+
2
1
e
+
f
=
0
a
+
b
+
c
+
d
+
e
+
f
=
0
2
5
9
a
+
2
5
2
1
b
+
2
5
4
9
c
−
5
3
d
−
5
7
e
+
f
=
0
Solve for a , b , c , d , e , f . Then verify that b 2 − 4 a c = 0 so that we know that it's a parabola, and not some other conic section.
The you plug in x = 2 4 0 2 8 9 to get a quadratic equation, and when you solve it, you'll find out that it has only one root, which is the answer.
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I love how you see and set the problem. It's obvious that since we have a solution ( 0 , 0 ) , f = 0 . Linear algebra can be tricky, but the best way to work the computation out is to compute the determinant for each variable.
Attempted to set matrix and determine its row echelon, but I end up with the different answer. Glad I got the right answer!
nice solution...........
I solve this question by taking the equation of the parabola as a x 2 + b y 2 + 2 . h x y + 2 g x + 2 f y + c = 0 , then it was same as yours. But i want to ask one thing that when you post such problems (hard ones,not conceptually but numerically), what do you expect from the solver to solve it using pen only or use of calculator is expected .? @Michael Mendrin @Calvin Lin
Let A = ( − 1 , 1 ) , B = ( 0 , 0 ) , C = ( 2 1 , 4 1 ) , D = ( 1 , 1 ) .
Let L ( A , B ) denote the equation of the line joining points A and B.
Then L ( A , D ) : y − 1 = 0 L ( B , C ) : x − 2 y = 0 L A , B ) : x + y = 0 , L ( C , D ) : 3 x − 2 y − 1 = 0
Now The parabola can be thought as a curve passing through the points of intersection of the curve ( y − 1 ) ( x − 2 y ) = 0 (which is combined equation of the lines L ( A , D ) ) and L ( B , C ) and the curve ( x + y ) ( 3 x − 2 y − 1 ) = 0 (which is combined equation of the lines L ( A , B ) and L ( C , D ) .
Hence equation of the parabola is of the form ( y − 1 ) ( x − 2 y ) + λ ( x + y ) ( 3 x − 2 y − 1 ) = 0
As parabola also passes through the point ( − 5 7 , − 5 3 ) , its coordinates satisfies the equation giving λ = − 2 5 1
Hence equation of the parabola is 2 5 ( y − 1 ) ( x − 2 y ) − ( x + y ) ( 3 x − 2 y − 1 ) = 0 i.e
x 2 + 1 6 y 2 − 8 x y + 8 x − 1 7 y = 0 whose vertex is at ( 2 4 0 2 8 9 , 9 6 0 7 9 9 )
So answer is 7 9 9 + 9 6 0 = 1 7 5 9
good solution
This implicit equation for a parabola passes through the 5 points
1 6 y 2 − 8 x y + x 2 − 1 7 y + 8 x = 0
At x = 2 4 0 2 8 9 , y is 9 6 0 7 9 9 , which is unique.
Hence, the answer is 1 7 5 9 .
This implicit equation does meet the condition for it to be a parabola, which is b 2 − 4 a c = 0 , since ( − 8 ) 2 − 4 ( 1 6 ) ( 1 ) = 0 .
Here's the graph of this parabola, with all the points plotted on it