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Find the slope of the normal to the curv...

Find the slope of the normal to the curve `x=1-asintheta,y=bcos^2theta`at `theta=pi/2`.

Text Solution

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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)`
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