If $h$ denotes the Arithmetic mean, $k$ denotes the geometric mean of the intercepts made on the axis by the lines passing through $(1,1)$, then $(h,k)$ lies on?
I don't even understand the question quite clearly although I tried something informal and ended up with $y^2=4x$. I don't know if it's correct. Can anyone please help me with a good solution and also correct my answer if it's wrong.
On
Let the line be $y=mx+c $ by putting $x,y=1,1$ thus we have $c=1-m $. From this we have line as $y=mx+1-m $ . From this the points of intercept are $(0,1-m), (\frac {m-1}{m},0) $ thus $h=\frac {\frac {m-1}{m}+1-m}{2}=\frac {-(m-1)^2}{2m} ,k=\sqrt {\frac {-(m-1)^2}{m}} $ this confirms $m<0$. We can see from here that $k^2=2h $ thus $y^2=2x $
Line passing through points $(1,1)$ is given by $$y-1=m(x-1)$$ on rearranging in the intercept form, we get $$\frac{y}{1-m}+\frac{x}{\frac{m-1}{m}}=1$$
Now let $a=1-m$ and $b=\frac{m-1}{m}$, here $a$ and $b$ are $x$ and $y$ intercepts of the line, respectively.
Given that $$h=\frac{a+b}{2}=-\frac{(m-1)^2}{2m}$$
And $$k^2=ab=-\frac{(m-1)^2}{m}$$
Using both the equations, you get $k^2=2h$ which is none other than the curve