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  • Equation of normal at the point with parameter  on the circle 

Equation of normal at the point with parameter \alpha=\frac{\pi}{3} on the circle x^2+y^2=4 is ?

Option: 1

y=\sqrt{3}x


Option: 2

y=-\sqrt{3}x


Option: 3

y=x


Option: 4

None of these


Answers (1)

best_answer

The point is (a\cos \theta, a\sin \theta) = (2\cos (\frac{\pi}{3}), 2\sin (\frac{\pi}{3}))= (1,\sqrt{3})

Centre of the circle is (0,0)

So the normal is the line passing through (0,0) and (1,\sqrt{3})

Equation of normal

\\(y-0)=\frac{\sqrt{3}-0}{1-0}(x-0)\\\\ y= \sqrt{3}x

Posted by

Divya Prakash Singh

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