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  • A movable parabola touches the x and the y-axes at (1,0) and (0,1). Then, the locus of the focus of the parabola isOption: 1 <img

A movable parabola touches the x and the y-axes at (1,0) and (0,1). Then, the locus of the focus of the parabola is

Option: 1

2 x^2-2 x+2 y^2-2 y+1=0


Option: 2

x^2-2 x+2 y^2-2 y+1=0


Option: 3

2 x^2-2 x+2 y^2+2 y+2=0


Option: 4

2 x^2+2 x-2 y^2-2 y-2=0


Answers (1)

best_answer

(a) Since, the x-axis and the y-axis are two perpendicular tangents to the parabola and both meet at the origin, the directrix passes through the origin.

Let y = mx be the directrix and (h,k) be the focus 

\begin{aligned} F A & =A M \\ \\\Rightarrow \quad \sqrt{(h-1)^2+k^2} & =\left|\frac{m}{\sqrt{1+m^2}}\right|\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i) \end{aligned}

\begin{aligned} F A & =A M \\ \\\Rightarrow \quad \sqrt{(h-1)^2+k^2} & =\left|\frac{m}{\sqrt{1+m^2}}\right|\: \: \: \: \: \: \: \: \: \: \: \: ....(ii) \end{aligned}

From Eqs. (i) and (ii), we get

(h-1)^2+h^2+h^2+(k-1)^2=1

\Rightarrow \quad \: \: \: \: \: \: \: \: 2 x^2-2 x+2 y^2-2 y+1=0

is the required locus.

Posted by

chirag

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