• # question_answer The equation of the parabola whose focus is the point (0, 0) and the tangent at the vertex is $x-y+1=0$is  [Orissa JEE 2002] A)            ${{x}^{2}}+{{y}^{2}}-2xy-4x+4y-4=0$ B)            ${{x}^{2}}+{{y}^{2}}-2xy+4x-4y-4=0$ C)            ${{x}^{2}}+{{y}^{2}}+2xy-4x+4y-4=0$ D)            ${{x}^{2}}+{{y}^{2}}+2xy-4x-4y+4=0$

Let focus is $S(0,\,0)$ and A is the vertex of parabola. Take any point Z such that AS = AZ. Given tangent at vertex is $x-y+1=0.$            Since directrix is parallel to the tangent at the vertex.                   $\therefore$ Equation of directrix is $x-y+\lambda =0$, where $\lambda$ is constant.            $\because$ A is midpoint of SZ,  $\therefore$ SZ = 2.SA            Þ$\frac{|0-0+\lambda |}{\sqrt{{{1}^{2}}+{{(-1)}^{2}}}}=2\times \frac{|0-0+1|}{\sqrt{{{1}^{2}}+{{(-1)}^{2}}}}$Þ$|\lambda |\,=2$  i.e. $\lambda =2$            $\because$ Directrix in this case always lies in IInd quadrant            \ $\lambda =2$            Hence equation of directrix is $x-y+2=0$            Now, P be any point on parabola            \ SP = PM Þ $S{{P}^{2}}=P{{M}^{2}}$            Þ ${{(x-0)}^{2}}+{{(y-0)}^{2}}={{\left( \frac{|x-y+2|}{\sqrt{2}} \right)}^{2}}$                    Þ ${{x}^{2}}+{{y}^{2}}+2xy-4x+4y-4=0.$