locus of point $P,$ If sum of slopes of normal is constant

Question

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

A supposed solution can be found here: https://doubtnut.com/question-answer/if-the-sum-of-the-slopes-of-the-normal-from-a-point-p-to-the-hyperbola-x-yc2-is-equal-to-lambdalambd-41041 but they also include the squares of complex slopes. — Intelligenti pauca, Feb 17 '20 at 20:38

lab bhattacharjee · Answer 1 · 2020-02-17T17:36:18.993

1

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

edited Feb 17 '20 at 17:36

answered Feb 17 '20 at 14:29

lab bhattacharjee

Your method would also compute in the sum complex values of the slopes, but this is probably not what is requested. – Intelligenti pauca Feb 17 '20 at 16:16
Moreover, the slope of the normal is $t^2$, not $1/t^2$. – Intelligenti pauca Feb 17 '20 at 16:47
@Aretino, thanks for the last feedback. With the given condition, is it possible to resolve the nature of the roots? – lab bhattacharjee Feb 17 '20 at 17:05
1

The coefficient of $u^3$ is $-h^2$, not $2hk$. – Intelligenti pauca Feb 17 '20 at 17:13

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