InterviewSolution
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InterviewSolution
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If the pair of straight lines `a x^2+2h x y+b y^2=0`is rotated about the origin through `90^0`, then find the equations in the new position. |
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Answer» Given equation of pair of straight lines is `ax^(2)+2hxy+by^(2)=0`. Let the component lines be `y=m_(1)xandy=m_(2)x`. `:.m_(1)+m_(2)=(-2h)/(b)andm_(1)m_(2)=(a)/(b)` Now , equations of perpendicular to above component lines are `y=-(1)/(m_(1))xandy=-(1)/(m_(2))x` or `m_(1)y+x=0andm_(2)y+x=0` Combined equation of above lines is `(m_(1)y+x)(m_(2)y+x)=0` or `m_(1)m_(2)y^(2)+xy(m_(1)+m_(2))+x^(2)=0` or `(a)/(b)y^(2)+xy((-2h)/(b))+x^(2)=0` or `bx^(2)-2hxy+ay^(2)=0` Alternative method : We can write the given equation of pair of straight lines as : `b((y)/(x))^(2)+2h((y)/(x))+a=0` The roots of this equations are `m_(1)andm_(2)` which are slopes of component lines . Now , we require equation whose roots are `-(1)/(m_(1))and-(1)/(m_(2))`. So, in above equation , replacing `(y)/(x)by-(x)/(y)`,we get `b(-(x)/(y))^(2)+2h(-(x)/(y))+a=0` or `bx^(2)-2hxy+ay^(2)=0` which is the required equation of pair of straight lines. |
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