Question

prove that the tangent to hyperbola xy=c^2 at point p(cp,c/p) is p^2y+x=2cp

Answer 1

\[xy=c^2\]\[y+xy'=0\]\[y'=-y/x\]\[y-y_0=y'(x_0)(x-x_0)=-\frac{y_0}{x_0}(x-x_0)\]\[y-c/p=-\frac{c/p}{cp}(x-cp)=-\frac{1}{p^2}x+c/p\]\[y+\frac{1}{p^2}x=2c/p\quad/\cdot p^2\]\[p^2y+x=2cp\]

Answer 2

where did you get the y+xy'=0?

Answer 3

Derivative of xy=c^2

Answer 4

There is another question by using the information above.the perpendicular line from the origin meets this tangent at N and meets the hyperbola again at Q and R.Prove that the N lies on the curve(x^2+y^2)^2=4c^2xy

Answer 5

\[\mbox{tangent:}\quad y=-\frac{1}{p^2}x+\frac{2c}{p}\]\[\mbox{perpendicular line:}\quad y=p^2x\]intersection of lines:\[p^2x=-\frac{1}{p^2}x+\frac{2c}{p}\quad\Rightarrow\quad x=\frac{2cp}{p^4+1}\quad,\quad y=\frac{2cp^3}{p^4+1}\]\[x^2+y^2=\frac{4c^2p^2}{(p^4+1)^2}+\frac{4c^2p^6}{(p^4+1)^2}=\frac{4c^2p^2(p^4+1)}{(p^4+1)^2}=\frac{4c^2p^2}{p^4+1}\]\[(x^2+y^2)^2=\frac{16c^4p^4}{(p^4+1)^2}=4c^2\frac{2cp}{p^4+1}\frac{2cp^3}{p^4+1}=4c^2xy\]

Answer 6

how did you get the perpendicular line y=p^2x? can you draw how it look like?

Answer 7

two perpendicular lines k1*k2=-1. If k1=-1/p^2, then k2=p^2

Answer 8

how did you get x and y?substitute from where?

Answer 9

Its ok.I found it.Btw,thanks.It helps a lot

Answer 10

system of two linear equations