On a point $(ct,c/t)$ on hyperbola $xy=c^2$ , a normal is drawn which intersects the hyperbola at $(ct',c/(t'))$ . Prove that $t^3t'=-1$ ?

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Answer 1 · 2017-07-28T07:28:57

Please see below.

The equation of hyperbola is $xy=c^2$ and point $(ct,c/t)$ lies on it.

Let us find the equation of the normal.

Equation of hyprbola can be written as $y=c^2/x$ and therefore slope of tangent is given by first derivative i.e. $(dy)/(dx)=-c^2/x^2$

hence slope of normal is given by $x^2/c^2$ and at $(ct,c/t)$ is $t^2$ and its equation is

$y=t^2(x-ct)+c/t$

or $xt^3-yt-ct^4+c=0$

As this passes through $(ct',c/(t'))$

$ct't^3-c/(t')t-ct^4+c=0$

or $ct^3(t'-t)+c/(t')(t'-t)=0$

as $t!=t'$ , $t-t'!=0$ and dividing by it we get

$ct^3=-c/(t')$

or $t^3t'=-1$

On a point (ct,c/t)(ct,c/t) on hyperbola xy=c^2xy=c^2, a normal is drawn which intersects the hyperbola at (ct',c/(t'))(ct',c/(t')). Prove that t^3t'=-1t^3t'=-1?