Saved Bookmarks
| 1. |
Find the equation of the normal to curve `x^2=4y`which passes through the point (1, 2). |
|
Answer» `h^2=4k` slope of normal`=-1/(dy/dx)=-2/h` equation of normal`(y-k)=-2/h(x-h)` `k=2+2/h(1-h)` `h^2/4=2+2/h(1-h)` h=2 k=1 equation of line (y-1)=-1(x-2) x+y=3 |
|