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  • Axis of a parabola is    and vertex and focus are at a distance  <img alt="\sqrt{2}" src="https

Axis of a parabola is  y=x  and vertex and focus are at a distance  \sqrt{2}  and  2\sqrt{2}  respectively from the origin. Then, equation of the parabola is

Option: 1

(x-y)^2=8(x+y-2)


Option: 2

(x+y)^2=2(x+y-2)


Option: 3

(x-y)^2=4(x+y-2)


Option: 4

(x+y)^2=2(x+y-2)


Answers (1)

best_answer

Since, distance of vertex from origin is  \sqrt{2}  and focus is 2 \sqrt{2} .

where, length of latusrectum 4a=4\sqrt2

∴ By definition of parabola 

PM^{2}=(4a)\, \, (PN)

where, PN is length of perpendicular upon x+y-2=0 , i.e. ,  tangent at vertex 

\begin{aligned} \Rightarrow & \frac{(x-y)^2}{2}=4 \sqrt{2}\left(\frac{x+y-2}{\sqrt{2}}\right) \\ \Rightarrow & (x-y)^2=8(x+y-2) \end{aligned}

Posted by

seema garhwal

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