Question
The axis of a parabola is along the line $$y = x$$ and the distances of its vertex and focus from origin are $$\sqrt 2 $$ and $$2\sqrt 2 $$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is -
A.
$${\left( {x + y} \right)^2} = \left( {x - y - 2} \right)$$
B.
$${\left( {x - y} \right)^2} = \left( {x + y - 2} \right)$$
C.
$${\left( {x - y} \right)^2} = 4\left( {x + y - 2} \right)$$
D.
$${\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right)$$
Answer :
$${\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right)$$
Solution :
Since, distance of vertex from origin is $$\sqrt 2 $$ and focus is $$2\sqrt 2 $$
$$\therefore $$ Vertex is $$\left( {1,\,1} \right)$$ and focus is $$\left( {2,\,2} \right),$$ directrix $$x+y=0$$
$$\therefore $$ Equation of parabola is
$$\eqalign{ & {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\frac{{x + y}}{{\sqrt 2 }}} \right)^2} \cr & \Rightarrow 2\left( {{x^2} - 4x + 4} \right) + 2\left( {{y^2} - 4y + 4} \right) = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 8\left( {x + y - 2} \right) \cr & \Rightarrow {\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right) \cr} $$
Since, distance of vertex from origin is $$\sqrt 2 $$ and focus is $$2\sqrt 2 $$
$$\therefore $$ Vertex is $$\left( {1,\,1} \right)$$ and focus is $$\left( {2,\,2} \right),$$ directrix $$x+y=0$$
$$\therefore $$ Equation of parabola is
$$\eqalign{ & {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\frac{{x + y}}{{\sqrt 2 }}} \right)^2} \cr & \Rightarrow 2\left( {{x^2} - 4x + 4} \right) + 2\left( {{y^2} - 4y + 4} \right) = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 8\left( {x + y - 2} \right) \cr & \Rightarrow {\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right) \cr} $$