Question

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. ?

Answer 1

Please see below.

Explanation:

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))`