Show that the locus of the point P (cp,c/p) is the curve xy=c^2. Show also that the tangent at point P has the equation yp^2+x=2cp. The vertical line from the origin meets the tangent at point T. Find the locus T as p varies. ?

The coordinates of point #P# are given in terms of parametric form, with parameter #p#, of the locus of hyperbola.

As #x=cp# and #y=c/p#, we eliminate parameter #p#

and get #xy=c^2#

To find the slope at #P(cp,c/p)#, we need to find the slope of tangent, which is nothing but first dervative #(dy)/(dx)#.

As derivative #(dy)/(dx)=((dy)/(dp))/((dx)/(dp))#

Now #(dy)/(dp)=-c/p^2# and #(dx)/(dp)=c#

and hence slope of tangent is #-1/p^2# and as it passes through #P(cp.c/p)#, equation of tangent is

#y-c/p=-1/p^2(x-cp)# or #yp^2-cp=-x+cp#

or #yp^2+x=2cp#

A line paerpendicular to this tangent and passing through #(0,0)# has equation #y-p^2x=0# (this is because a line perpendicular to #ax+by+c=0# is of the type #bx-ay+k=0#, but as it passes through #(0,0)#, constant term #k=0#).

We can find the point of intersection of tangent #yp^2+x=2cp# and this line #y=p^2x#, by solving these equations as simultaneous equations and as #y=p^2x#, we get

#p^4x+x=cp# or #x=(cp)/(1+p^4)#

and #y=(cp^3)/(1+p^4)#,

which gives parametric equation for locus of point as #T((cp)/(1+p^4),(cp^3)/(1+p^4))#