A curve is given by the parametric equations x=t,y=#1/t#. (pls see details below). ?

(a) Find the equation of line passing through the points (s,#1/s#) and (t,#1/t#).
(b) Deduce the equation of tangent to the curve at the point (t,#1/t#).
(c)Find the equation of normal to the curve at the point (t,#1/t#).
If this normal cuts the curve again, find the coordinates of point of intersection

1 Answer
Jun 25, 2018

Line through...: #tx + t^2 s y = t^2-s^2#

Tangent: #x +t^2 y=2t#

Normal: #t^3 x - t y = t^4 -1#

Other intersection: #(-1/t^3,-t^3)#

Explanation:

Fun. We have #(x,y)=(t,1/t).# To deparameterize (not that we have to) that's #x=t# so this is just the standard hyperbola #y=1/x.#

The line through #(s.1/s)# and #(t,1/t)# is

# (y-1/s)(t-s)=(x-s)(1/t - 1/s)#

Multiplying both sides by #st#,

#t(sy-1)(t-s)=(x-s)(s-t)#

#t(sy-1)=-(x-s)#

#x + sty = s+t#

We get the tangent line as #s to t#

#x +t^2 y=2t#

For the normal line we swap the coefficients on #x# and #y#, negating one, then use the point #(t,1/t)# for the constant:

#t^2 x - y = t^2(t)-1/t = t^3 -1/t#

#t^3 x - t y = t^4 -1#

We plug in #y=1/x# to find where the normal meets the curve

#t^3 x - t/x = t^4 -1#

Multiply by #x# to clear the denominator.

#t^3 x^2 - (t^4-1)x - t = 0#

We know #x=t# is one root so we can factor:

#(x-t)(t^3 x + 1) = 0#

#x=t# we know about; the other point is #(-1/t^3,-t^3)#

Let's plot these at #t=2# so #(x,y)=(2,1/2)# with the Socratic #cancel{"crasher"}# grapher.

Plot #0=(y-1/x)(x +2^2 y - 2(2))(2^3 x - 2 y - (2^4 -1))( (x- (-1/2^3) )^2+(y - (-2)^3)^2-.1^2) #

graph{0=(y-1/x)(x +2^2 y - 2(2))(2^3 x - 2 y - (2^4 -1))( (x- (-1/2^3) )^2+(y - (-2)^3)^2-.2^2) [-20, 20, -10, 10]}