- #1

chwala

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- Homework Statement
- See attached

- Relevant Equations
- Area of triangle

I am still looking at this question. One thing that i know is that the

distance ##AB=\dfrac {(λ+1)\sqrt {2λ^2-2λ+1}}{λ^2}##

distance ##OA=\sqrt 2##

distance ##OB=\dfrac{\sqrt{λ^2+1}}{λ^2}##

Perpendicular distance from point ##B## to the line ##OA=\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##

Therefore, the area of the triangle is given by;

##\dfrac {1}{2} ×OA ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##

##=\dfrac {1}{2} ×\sqrt 2 ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##

##=\dfrac {1}{2} ×\sqrt 2 ×\sqrt 2 ×\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##

##=\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##

Now as i look at this am in doubt on how they arrived at the height ##h=\dfrac {1}{\sqrt {2λ^2-2λ+1}}##...did they assume that ##O## will meet the line ##AB## at its mid point? If so, then the solution given may not be correct... otherwise, i do not see any other way...am still trying to figure this out.

Are they also assuming that the line ##AB## is parallel to the ##x- axis##? Point ##B## could be anywhere on the first quadrant since we are told that ##λ>0##.

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