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Find the area of the triangle formed by the positive x-axis and the tangent and normal to the curve x2+y2=9 at (2,√5)


2 years ago

Susmita
425 Points
							Given x2+y2=9. This is circle of radius 3 with centre at origin (0,0).For equation of tangent,we have to calcute dy/dx.From the given equationy2=9-x2Differentiate both sides wrt x2y(dy/dx)=-2xor,dy/dx=-x/ydy/dx at the point (2,root 5) is$\frac{dy}{dx}=\frac{-x}{y}=\frac{-2}{\sqrt{5}}$So equation of tangentY-y=m(X-x)$Or,Y-\sqrt{5} = \frac{-2}{\sqrt{5}}(X-2)$$Or,Y= \frac{-2x}{\sqrt{5}} +\frac{9}{\sqrt{5}}$This tangent intersect the positive x axis at a point where Y=0,ie,X=9/2=4.5As the curve is a circle,so the normal at (2,root5) will be the radius drawn from (2,root5) to centre (0,0).Please draw a triangle with vertices at (0,0),(4.5,0),(2,root5).Now drop a straight line from (2,root5) to (2,0).This line will divide the triangle into two sections.For the 1st section,the base length is 2 and height is root5.So area is (1/2)*2*root5=root5.For the second segment,base length is (4.5-2)=2.5 and height is root5.So area is (1/2)*2.5*root5=9root5/4.Add up these two area and that is the final answer.Ask me if you face any difficulty with my instruction.If you are helped please approve the answer.

2 years ago
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