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) \[{{x}^{2}}+{{y}^{2}}+2xy-4x-4y+4=0\]
Correct Answer: C
Solution :
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 \[{{x}^{2}}+{{y}^{2}}+2xy-4x+4y-4=0.\]You need to login to perform this action.
You will be redirected in
3 sec