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

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.

• What does "AB, AP, AC are in HP" mean? Especially what is "HP" the abbreviation of? – Ingix Feb 13 '19 at 17:42
• The distances of AB, AP and AC are in Harmonic progression. Here in India we use these shorter notations. – jayant98 Feb 13 '19 at 17:47
• Please add your results for $B$ and $C$. – Intelligenti pauca Feb 13 '19 at 18:04
• @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. – jayant98 Feb 13 '19 at 18:29

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.

Now the problem should not be difficult to solve.

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