How to find the minimum distance from origin to locus of P?

Question

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.

Answer 1

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 $B=C=P=D$:

Answer 2

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.

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.

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Now the problem should not be difficult to solve.