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# 2nd Repost

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The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$. For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\leq 1000$?

Jan 1, 2021

### 6+0 Answers

#1
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1250n+251250n+25 to 1250n+12251250n+1225. It turns out that only when n=0n=0 or 11 are the criteria satisfied, and in each case there are 2020 of the pairs. So 40 is the answer.

For more in debt answer just ask meh

-wolfie

(If wrong im sorry my brain hurts )

Jan 1, 2021
edited by wolfiechan  Jan 1, 2021
edited by wolfiechan  Jan 1, 2021
#3
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And

congrats on getting a amc 12 problem correct, the hardest problem on the test was correct,y answered by you. You are surely underrated

Jan 1, 2021
#4
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Concave up parabola

Focus(0,0)  A point (4,3)

Distance of the point to the focus is 5

Distance of (4,3) to the directrix is 5

So the directrix is  y=-2

So the vertex is (0,-1)

Equation is

$$(x-0)^2=4a(y+1)\\ x^2=4(y+1)\\ so\\ y=\frac{x^2}{4}-1=\frac{x^2-4}{4}$$

$$4x+3y \le1000\\ 4x+3(\frac{x^2-4}{4})\le1000\\ 16x+3x^2-12\le4000\\ 16x+3x^2-12\le4000\\ 3x^2+16x-4012\le 0\\ find\;\;roots\\ x=\frac{-16\pm\sqrt{256+12*4012}}{6}\\ x=\frac{-16\pm 220}{6}\\ x=-39.\dot3\;\;or\;\;\;x=34$$

So this will be true for all integer values of x from -39 to  34  =  34- -39+1 = 74 integer x values.

BUT how many of these integer x values will have integer y values.

Only those where x^2 is divisible by 4.  If x^2 is divisible by 4 then x is divisible by 2. So x must be even.

So how many even numbers are there from -38 to 34 inclusive?   17+19+1 = 37

So 37 coordinate pairs will make this statement true.

Here is the graph. Jan 1, 2021
#5
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Hello Melody,

Here's the answer explained by AoPS.

https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_25

It seems like the answer is 40.

=^._.^=

catmg  Jan 1, 2021
#6
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Ok, thanks catmg.

I misread the question.

Mine is based on the parabola going through the points  (4,3)  and  (-4,+3)

Melody  Jan 2, 2021
edited by Melody  Jan 2, 2021