InterviewSolution
Saved Bookmarks
| 1. |
Find the equation of tangents to hyperbola `x^(2)-y^(2)-4x-2y=0` having slope 2. |
|
Answer» We have hyperbola `x^(2)-y^(2)-4x-2y=0` `"or "(x-2)^(2)-(y+1)^(2)=3` `"or "((x-2)^(2))/(3)-((y+1)^(2))/(3)=1" (1)"` Equation of tangents of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` having slope m is given by `y=mxpm sqrt(a^(2)m^(2)-b^(2))` So, equation of tangents to hyperbola (1) having slope 2 is given by `y+1=2(x-2)pmsqrt(3xx4-3)` `"or "y+1=2x-4pm3` Therefore, the equations of tangents to given hyperbola are `2x-y-2=0` and `2x-y-8=0.` |
|