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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 $|4x+3y|\le 1000$

$A:38 , B:40 C:42 D:44 E:46$

My idea:

maybe this $(x,y)$ such $$\sqrt{x^2+y^2}=\dfrac{|3x-4y-25|}{5}$$ $$25(x^2+y^2)=9x^2+16y^2-24xy-150x+200y+625$$ $$\Longrightarrow 16x^2+9y^2+24xy+150x-200y=625$$

Then I can't, Thank you very much

dcstup
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math110
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1 Answers1

1

$B. 40$

You can solve $16x^2+9y^2+24xy+150x-200y-625=0$ over the integers on WolframAlpha, and then use the parametric expression of $x$ and $y$ to find a expression of $4x+3y$.

The expression you'll find is $4x+3y=1250n+50p-25$, $n$ being a integer and $p$ being a positive integer that is less than $\frac{1250}{50}$.

Then, you can build a simple excel file to exhaust the answers from $1250n+25$ to $1250n+1225$. It turns out that only when $n=0$ or $1$ are the criteria satisfied, and in each case there are $20$ of the pairs. So $B. 40$ is the answer.

This is taken from AMC 12A 2014 if I'm not wrong, so the solution should not require computer.I'm not sure about how to work it out without computer though. The exhaustion can be easily done by hand so the only step that requires thinking is to find the general expression of integer solutions. You can go to Art of Problem Solving forum for a Math Jam of AMC 10A/12A discussion.

dcstup
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