find the area of the triangle formed by the positive x axis and the tangent and normal to the curve x^2 + y^2 =9 at(2,√5).
find the area of the triangle formed by the positive x axis and the tangent and normal to the curve x^2 + y^2 =9 at(2,√5).
The curve you will get from that EQ. Is a a circle from centre with radius 3.
The equation of tangent from the point is
2x+(√5)y=9
and the equation of normal is
(√5)x =2y (passing through origin)
Now we know two points of the triangle.. i.e (0,0) and (2,√5) and the third point would be the POI of x axis(y=0) and the tangent.
Puting y=0 in the equation of tangent, we get x=9/2.
So now we have all three points. You can use these co-ordinates to find the area of triangle (by various formulas) or just use the simple formula of b*h/2
For the triangle formed by point A(0,0), B(9/2,0) and C(2,√5). let the base be AB. Also the y co-ordinate of C would then be the height.
Applying the formula for area of triangle
Area=b*h/3
=(9/2)*(√5)*(1/2)
=9√5/4
The curve you will get from that EQ. Is a a circle from centre with radius 3.
The equation of tangent from the point is
2x+(√5)y=9
and the equation of normal is
(√5)x =2y (passing through origin)
Now we know two points of the triangle.. i.e (0,0) and (2,√5) and the third point would be the POI of x axis(y=0) and the tangent.
Puting y=0 in the equation of tangent, we get x=9/2.
So now we have all three points. You can use these co-ordinates to find the area of triangle (by various formulas) or just use the simple formula of b*h/2
For the triangle formed by point A(0,0), B(9/2,0) and C(2,√5). let the base be AB. Also the y co-ordinate of C would then be the height.
Applying the formula for area of triangle
Area=b*h/3
=(9/2)*(√5)*(1/2)
=9√5/4
