A variable straight line passes through the fixed point $A(6,1)$ and meets the ellipse $x^2 + 2y^2 = 2$ at points $B$ and $C$. If $P$ is a point such that the lenghts $AB, AP, AC$ are in HP (harmonic progression), find the locus of $P$ in the $X-Y$ plane.
I ruled out the approach of assuming the line to be $y=mx+c$ and finding a relation between $m$ and $c$ by putting $x=6$ and $y=1$, then solving it with the ellipse to find $B$ and $C$ in very tedious terms of m and so on and so forth. I think determining the line in terms of polar parameters $r$ and $\theta$ is the best approach. Let for some $r$ and $\theta$, $(6+rcos\theta, 1+rsin\theta)$ lie on the ellipse.
Solving with these parameters gives you a quadratic in $r$, whose roots are lengths $AB$ and $AC$. Using a bit of Vieta and applying condition of HP, you get a condition for AP in terms of $\theta$. $AP$ is the length of the vector from point A to P, but the parameter still exists and I am not able to figure out how to eliminate $\theta$. Is there a better approach, or am I missing something?
What I did:
Consider a line passing through $A(6,1)$ with slope as $tan\theta$. Hence, any point on the line can be written in terms of its distance $r$ from point $A$ as $(6+rcos\theta,1+rsin\theta)$. Let the line intersect the ellipse for some value of $r$. Solving the line with the ellipse, we get:
$(6+rcos\theta)^2 + 2(1+rsin\theta)^2 = 2$
On simplifying, we get the equation,
$(1+sin^2\theta)r^2 + 4(sin\theta + 3 cos\theta)r + 36 = 0$
Let $r_1,r_2$ be the roots of the equation.
Therefore, $r_1+r_2=-4(sin\theta+3cos\theta)/(1+sin^2\theta)$
and $r_1r_2 = 36/(1+sin^2\theta)$
According to condition of HP, $AP = 2(AB)(AC)/(AB + AC)$
Therefore, $AP = -18/(sin\theta + 3cos\theta)$
But that is all. Substituting $P$ as $(h,k)$ here does not help.
I will be continuing from where I left off in the method above.
Let $r'= AP=-18/(3cos\theta+sin\theta)$ Therefore, distance of point $P$ from $A$ is $r'$.
Let locus of P be defined by coordinates $(h,k)$.
Therefore, $h=6+r'cos\theta$ $\rightarrow$ $h-6=-18cos\theta/(3cos\theta+sin\theta)$ $\rightarrow$ $-(h-6)/6=3cos\theta/(3cos\theta+sin\theta)$
Similarly, $-(k-1)/18=sin\theta/(3cos\theta+sin\theta)$
Therefore, $-(h-6)/6 - (k-1)/18=1$ $\rightarrow 3h+k=1$
Hence, locus of $P$ is $3h+k=1$.