InterviewSolution
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InterviewSolution
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The square of the distance between the origin and the point of intersection of the lines given by `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`, isA. `(c(a+b)+f^(2)+g^(2))/(ab-h^(2))`B. `(c(a+b)-f^(2)-g^(2))/(h^(2)-ab)`C. `(c(a+b)-f^(2)-g^(2))/(ab-h^(2))`D. none of these |
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Answer» Correct Answer - C Let `f(x,y)=ax^(2)+2hxy+by^(2)+2gx+2fy+c.` The point of intersection of the lines given by `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0` is obtained by solving `(delf)/(delx)=0` and `(delf)/(dely)=0` i.e. `ax+hy+g=0 and hx+by+f=0` The coordinates of the point of intersection are `P((hf-bg)/(ab-h^(2)),(hg-af)/(ab-h^(2)))` `therefore" Distance between the origin and the point P is given by "` `OP=sqrt(((hf-bg)^(2)+(hg-af)^(2))/((ab-h^(2))^(2)))` `=sqrt((f^(2)(h^(2)+a^(2))+g^(2)(h^(2)+b^(2))-2fgh(a+b))/((ab-h^(2))_(2)))` `=sqrt((f^(2)(h^(2)+a)+g^(2)(h^(2)+b^(2))-(a+b)(af^(2)+bg^(2)+ch^(2)-abc))/((ab-h^(2))^(2)))` `" "[because abc+2fgh-af^(2)-hg^(2)-ch^(2)=0]` `=sqrt((ac(ab-h^(2))+bc(ab-h^(2))-f^(2)(ab-h^(2))-g^(2)(ab-h^(2)))/((ab-h^(2))^(2)))` `=sqrt((c(a+b)-f^(2)-g^(2))/((ab-h^(2))))` Hence, `OP^(2)=(c(a+b)-f^(2)-g^(2))/((ab-h^(2)))` |
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