Using integration find the area of triangle formed by -ve x axis,the tanget and normal to the circle x
^2+y^2=9 at the point (-1,2* 2^1/2)

Asked by Balbir | 29th Nov, 2017, 10:13: PM

Expert Answer:

begin mathsize 16px style tangent space to space std space circle space is space given space by space
straight y equals mx plus-or-minus straight r square root of 1 plus straight m squared end root
now space straight r equals 3 comma space straight y equals 2 square root of 2 comma space straight x equals negative 1
we space get
8 straight m squared minus 4 square root of 2 straight m plus 1 equals 0
straight m equals fraction numerator square root of 2 over denominator 4 end fraction
now space you space can space find space the space ans space using space geometry
so space equation space of space line space is
straight y equals fraction numerator square root of 2 over denominator 4 end fraction straight x plus 8 square root of 1 plus open parentheses fraction numerator square root of 2 over denominator 4 end fraction close parentheses squared end root space consider space plus straight r space since space 3 rd space quadrant
y equals fraction numerator square root of 2 over denominator 4 end fraction straight x plus 6 square root of 2
end style
begin mathsize 16px style finding space the space straight x space intercept space and space straight y space intercept space
and space applying space 1 half cross times straight b cross times straight h
OR
thru space integration
integral subscript negative 24 end subscript superscript 0 open parentheses fraction numerator square root of 2 over denominator 4 end fraction straight x plus 6 square root of 2 close parentheses end style.

Answered by Arun | 30th Nov, 2017, 01:14: PM