# 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)

Susmita
425 Points
5 years ago
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 equation
y2=9-x2
Differentiate both sides wrt x
2y(dy/dx)=-2x
or,dy/dx=-x/y
dy/dx at the point (2,root 5) is
$\frac{dy}{dx}=\frac{-x}{y}=\frac{-2}{\sqrt{5}}$
So equation of tangent
Y-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=4.5
As 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.