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find the locus of points M, the difference between the squares of the distances from which to two given points A and B is equal to a given value C, At which C does the problem have a solution?

Anyone know how to solve this?

Amarie Ruiz
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1 Answers1

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This locus is a line perpendicular to $AB$. For proving it we'll prove the following:

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In Figure1, $AB$ is perpendicular to $CD$ iff $AC^2-AD^2=BC^2-BD^2$

Proof: Assume that $H_a,H_b$ are the feet of altitudes from $A,B$ to $CD$, respectively. By pythagorean theorem,
$$AC^2-AD^2=AH_a^2+CH_a^2-AH_a^2-DH_a^2=CH_a^2-DH_a^2$$
Similarly, $BC^2-BD^2=CH_b^2-DH_b^2$. Therefore it suffices to prove that $H_a\equiv H_b$ iff $CH_a^2-DH_a^2=CH_b^2-DH_b^2$ and this is trivial since:
$$CH_a^2-DH_a^2=CH_b^2-DH_b^2 \implies (CH_a+DH_a)(CH_a-DH_a)=(CH_b+DH_b)(CH_b-DH_b)$$
But since $CH_a+DH_a=CH_b+DH_b$,
$$CH_a-DH_a=CH_b-DH_b$$
But since sum of them are also equal, $CH_a=CH_b$ and hence the result.

Intelligenti pauca
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Aryan
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  • So how would I apply what you stated to my problem? Thank you! – Amarie Ruiz Jan 25 '21 at 21:07
  • Consider the point on $AB$ like $P$ such that $AP^2-BP^2$ is equal to the fixed value you're considering. Then due to this theorem, any other point like $P'$ has the property that $AP'2-BP'^2$ is equal to the same fixed value iff $PP'$ is perpendicular to $AB$. So the desired locus is always a line perpendicular to$AB$ and the location of this perpendicular line depends on the fixed value. – Aryan Jan 25 '21 at 21:30