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  • Show that the normal at any point q to the curve x = a cos theta + a theta sin theta , y = a sin theta - a theta cos theta

5. Show that the normal at any point \theta to the curve x = a \cos \theta + a \theta \sin \theta , y = a \sin \theta - a\theta \cos\theta

is at a constant distance from the origin.

Answers (1)

best_answer

We know that the  slope of tangent at any point is given by  \frac{dy}{dx}
Given equations are
x = a \cos \theta + a \theta \sin \theta , y = a \sin \theta - a\theta \cos\theta
\frac{dx}{d\theta} = -a\sin \theta + a\sin \theta -a\theta\cos \theta = -a\theta\cos \theta
\frac{dy}{d\theta} =a\cos \theta -a\cos \theta +a\theta (-\sin \theta) = -a\theta\sin \theta
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a\theta\sin \theta}{-a\theta \cos \theta} = \tan \theta
We know that 
Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{\tan \theta}
equation of normal with given points and slope
y_2-y_1=m(x_2-x_1)\\ y - a\sin \theta + a\theta\cos\theta = \frac{-1}{\tan \theta}(x-a\cos\theta-a\theta\sin\theta)\\ y\sin\theta - a\sin^2 \theta + a\theta\cos\theta\sin\theta = -x\cos\theta+a\cos^2\theta+a\theta\sin\theta\cos\theta\\ y\sin\theta + x\cos\theta = a
Hence, the equation of normal is  y\sin\theta + x\cos\theta = a
Now perpendicular distance of normal from the origin (0,0) is
D = \frac{|(0)\sin\theta+(0)\cos\theta-a|}{\sqrt{\sin^2\theta+\cos^2\theta}} = |-a| = a = \ constant \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\ (\because \sin^2x+\cos^2x=1)
Hence, by this, we can say that 

the normal at any point \theta to the curve x = a \cos \theta + a \theta \sin \theta , y = a \sin \theta - a\theta \cos\theta

is at a constant distance from the origin

Posted by

Gautam harsolia

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