Question

Y^2+x^2p^2-2xyp=4/p^2find GS and SS
Where GS=y^2-2cxy+c^2(x^2-1)=m^2
SS= y^2+m^2=m^2 answers to be got

Answer 1

Given equation is incorrect

Correct equation is,

$y^2+x^2p^2-2xyp=4$

$(y-xp)^2=2^2$

$\to y-xp=2$

$\to y-x\dfrac{dy}{dx}=2$

$\to y-2=x\dfrac{dy}{dx}$

$\to \dfrac{dy}{y-2}=\dfrac{dx}{x}$

Integrating Both the sides

$\to \int \dfrac{dy}{y-2}=\int\dfrac{dx}{x}$ '

$\therefore log(y-2)=logx+logc$

$\to y-2=xc$

$\to y=xc+2$

For general solution ,

as y=xc+2

or y-xc=2

Squaring on both sides

$\to (y-xc)^2=2^2$

$\to y^2+x^2c^2-2xyc=4\\$

Subtract $c^2$ on both sides

$\to y^2+x^2c^2-2xyc-c^2=4-c^2\\ \to y^2-2xyc+c^2(x^2-1)=m^2$ ( where $m^2=4-c^2$ )

For SS

Let calculate value of constant at (0,0)

$\to 0+c^2(-1)=m^2\\ c^2=-m^2$

So SS is

$\to y-0+c^2(-1)=m^2\\ \to y-c^2=m^2\\ \to y+m^2=m^2$