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A straight line through $A(6,8)$ meets the curve $ 2x^2+y^2=2$ at $B$ and $C$. $P$ is such a point on $BC$ that the distances $AB, AP, AC$ are in Harmonic Progression. Find minimum distance from origin to the locus of $P$.

$**Attempt**$

Took $y=mx+c$ and put $(6,8)$ in it to get $8=6m+c$ as one of the relation and then this line cuts the curve at two points so solved these two and got $B$ and $C$. Also I used the 1st relation I got in this process. But from here I don't know what should I do?

Any hints or suggestions?

Edit: For B and C

$$x=\frac{{12m^2-16m}(+or-)\sqrt{-280m^2+768m-496}}{4+2m^2}$$ and putting these two values in $y=mx+8-6m$ we get B and C.

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    $\begingroup$ What does "AB, AP, AC are in HP" mean? Especially what is "HP" the abbreviation of? $\endgroup$ – Ingix Feb 13 '19 at 17:42
  • $\begingroup$ The distances of AB, AP and AC are in Harmonic progression. Here in India we use these shorter notations. $\endgroup$ – jayant98 Feb 13 '19 at 17:47
  • $\begingroup$ Please add your results for $B$ and $C$. $\endgroup$ – Intelligenti pauca Feb 13 '19 at 18:04
  • $\begingroup$ @Aretino $x=\frac{{12m^2-16m}(+or-)\sqrt{-280m^2+768m-496}}{4+2m^2}$ and putting these two values in $y=mx+8-6m$ we get B and C. $\endgroup$ – jayant98 Feb 13 '19 at 18:29
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Here is a general result.

If $P$ is such that $AB$, $AP$ and $AC$ make harmonic progression, then $P$ and $A$ are harmonic conjugate with respect to a given ellipse. That is, $P$ is on a polar line for $A$ with respect to a given ellipse.

enter image description here

Proof: We have $$ AP = {2\over {1\over AB}+{1\over AC}} \implies AP\cdot(AB+AC)= 2AB\cdot AC$$

Using notation on picture we get $$b\cdot (a+b+c) = a\cdot c\implies PB\cdot AC = AB\cdot PC$$

so we have $${\vec{BA}\over \vec{AC}}:{\vec{BP}\over \vec{PC}}=-1$$ and thus the claim.


Now the problem should not be difficult to solve.

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Let $D$ end $E$ be a touching points of tangents from $A$ to ellipse.

Playing in Geogebra, we can see that the maximum is achieved iff $B=C=P=E$ and minimum iff $B=C=P=D$:

enter image description here

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  • $\begingroup$ D and E are which points? P is on the BC line as per the question. $\endgroup$ – jayant98 Feb 13 '19 at 19:01

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