• # question_answer The equation of the common tangent to the curves ${{y}^{2}}=8x$ and $xy=-1$ is        [IIT Screening 2002] A)            $3y=9x+2$                                B)            $y=2x+1$ C)            $2y=x+8$                                  D)            $y=x+2$

Any point on ${{y}^{2}}=8x$ is $(2{{t}^{2}},\,4t)$ where the tangent is $yt=x+2{{t}^{2}}.$            Solving it with $xy=-1,$ $y(yt-2{{t}^{2}})=-1$             or $t{{y}^{2}}-2{{t}^{2}}y+1=0$.            For common tangent, it should have equal roots.                   \ $4{{t}^{4}}-4t=0$ Þ $t=0,\,1$.                      $\therefore$ The common tangent is $y=x+2$, (when $t=0$, it is $x=0$ which can touch $xy=-1$ at infinity only).