 
                 
                InterviewSolution
 Saved Bookmarks
    				| 1. | If the roots of the equation `(x_(1)^(2)-a^2)m^2-2x_1y_1m+y_(1)^(2)+b^2=0(agtb)` are the slopes of two perpendicular lies intersecting at `P(x_1,y_1)`, then the locus of P isA. `x^2+y^2=a^2+b^2`B. `x^2+y^2=a^2-b^2`C. `x^2-y^2=a^2+b^2`D. `x^2-y^2=a^2-b^2` | 
| Answer» Correct Answer - B Equation `(x_1^(2)-a^2)m^2-2x_1y_1m+y_1^(2)+b^2=0` has roots `m_1` and `m_2` `therefore m_1m_2=(y_(1)^(2)+b^2)/(x_(1)^(2)-a^2)=-1` (Given) `therefore x^2+y^2=a^2-b^2` | |