Saved Bookmarks
| 1. |
Find the slope of the normal to the curve `x=1-asintheta,y=bcos^2theta`at `theta=pi/2`. |
|
Answer» Let `m` is the slope of tangent of the curve and `n` is the slope of normal to the curve. Then, `m*n = -1=> n = -1/m`. Now, equation of the curve, `x = 1-asintheta , y = bcos^2theta` `:. m = dy/dx = (dy/(d theta))/(dx/( d theta)) = (b(2cos theta)(-sintheta))/(-acostheta)` `=> m = (2b)/a sin theta` `:. n = -1/m = -a(2bsintheta)` At, `theta = pi/2`, `n = -a/(2b)` So, solope of normal to the given curve at `theta = pi/2` is `-a/(2b).` |
|