InterviewSolution
Saved Bookmarks
| 1. |
Find the equation of the tangent line to the curve `y=sqrt(5x-3)-2`which is parallel to the line `4x-2y+3=0` |
|
Answer» slope = `m= -4/(-2) = 2` `dy/dx|_p = 2` `=> 1/(2sqrt(5x-3)) xx 5 = 2` `x = 73/80` putting it in the curve eqn `y= -3/4 p(75/80, -3/4)` `m=2` eqn of tangent `y-(-3/4) = 2(x-73/80)` Answer |
|