InterviewSolution
Saved Bookmarks
| 1. |
the value of `lambda` for which the line `2x-8/3lambday=-3` is a normal to the conic `x^2+y^2/4=1` is: |
|
Answer» Given normal to ellipse is `2x-(8)/(3)lamday=-3` or `y=((3)/(4lambda))x+((9)/(8lambda))` Here, `m=(3)/(4lambda)=(9)/(8lambda)` Now, condition for line y=mx+c to be normed to ellipse is `C=+-((a^(2)-b^(2))m)/(sqrt(a^(2)+b^(2)m^(2)))` `rArr(9)/(8lambda)=+-(3(3)/(4lambda))/(sqrt(1+4(9)/(16lambda^(2))))` `rArr 4lambda^(2)+9=16lambda^(2)` `rArr4lambda^(2)=3` `rArr=+-(sqrt(3))/(2)` Thus, normals are `2x+-(4)/(sqrt(3))y=-3` |
|