If the sum of the slopes of the normals from a point $P$ on hyperbola $xy=c^2$ is constant $k(k>0),$ Then the locus of point $P$ is
what i try
Let coordinates of point be $\displaystyle P\bigg(ct,\frac{c}{t}\bigg)$
$$xy'+y=0\Rightarrow y'(m_{t})=-\frac{y}{x}\bigg|_{(ct,\frac{c}{t})}=-\frac{1}{t^2}$$
slope of normal $$(m_{n})=t^2$$
equation of normal is $$y-\frac{c}{t}=t^2(x-ct)\Rightarrow yt-c=t^3(x-ct)$$
$$xt^3-ct^4=yt-c$$
$$ct^4-xt^3+yt-c=0$$
how do i solve it help me please
The equation of normal at any point on the hyperbola,
$$ct^4-xt^3+yt-c=0$$
Now if this passes through $P(h,k)$
$$ct^4-ht^3+kt-c=0$$
Clearly there are four possible values of $t$
As the slope of normal is $t^2=u$(say)
$$c^2(t^4-1)^2=t^2(ht^2-k)^2$$
$$c^2(u^2-1)^2=u(hu-k)^2$$
$$c^2u^4+2h^2k^2u^3+\cdots=0$$
Use Vieta's formula
$-\dfrac{2h^2k^2}{c^2}=$constant