InterviewSolution
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InterviewSolution
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If `x^2+2h x y+y^2=0`represents the equation of the straight lines through the origin whichmake an angle `alpha`with the straight line `y+x=0``s e c2alpha=h``cosalpha``=sqrt(((1+h))/((2h)))``2sinalpha``=sqrt(((1+h))/h)`(d) `cotalpha``=sqrt(((1+h))/((h-1)))`A. `sec2alpha=h`B. `cosalpha=sqrt((1+h)//(2h))`C. `2sinalpha=sqrt((1+h)//h)`D. `cot alpha=sqrt((h+1)//(h-1))` |
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Answer» Correct Answer - 1,2,4 Let the equation of the lines given by `x^(2)+2hxy+y^(2)=0` be `y=m_(1)xandy=m_(2)x`. Since these make an angle `alpha` with `y+x=0` whose slope is -1, we have `(m_(1)+1)/(1-m_(1))=tanalpha=(-1-m_(2))/(1-m_(2))` or `m_(1)+m_(2)=((tanalpha-1)^(2)+(tanalpha+1)^(2))/(tan^(2)alpha-1)` `=(-sec^(2)alphaxxcos^(2)alpha)/(cos2alpha)` `:.-2sec2alpha=-2h` or `sec2alpha=h` or `cos2alpha=(1)/(h)or2cos^(2)alpha-1=(1)/(h)` or `cosalpha=sqrt((1+h)/(2h))andcotalpha=sqrt((h+1)/(h-1))` |
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