The line $$x+y=a$$ meets the axis of x and y at A and B respectively a triangle AMN is inscribed in the triangle OAB, O being the origin, with right angle at N,M and N lie respectively on OB and AB. If the area of the triangle AMN is $$3/8$$ of the area of the triangle OAB, then $$AN/BN$$ is equal to

Question

The  line $$x+y=a$$     meets the axis of x and y at A and B respectively a  triangle AMN is inscribed in the triangle OAB, O being the origin, with right  angle at N,M and N lie respectively on OB and AB. If the area of the triangle  AMN is $$3/8$$ of the area of the triangle OAB, then $$AN/BN$$ is equal to

Infinity Learn · Accepted Answer

Let  $$ANBN=$$λ       Then,  coordinate of N is (a1+$$λ,aλ$$1+$$λ$$)     ∵      Slope  of  $$AB=$$−$$1$$                      ∴      Slope  of  $$MN=1$$     ∴     equation of MN isy−aλ$$1+$$λ$$=x$$−$$a1+$$λ                           ⇒                  x−$$y=a(1$$−λ$$1+$$λ$$)$$     So,  the coordinates of M are (0$$,$$a($$λ−$$1$$λ$$+1))     Therefore,  area of Δ$$AMN=38area$$  of  ΔOAB                    ⇒       $$12$$.AN.$$MN=38.12a$$.a                  ⇒       $$12$$.|aλ$$21+$$λ.$$a21+$$λ|$$=38.12a$$.a     ⇒       $$a2$$λ$$(1+$$λ$$)2=38.12a2$$     ∴           λ$$=3$$  or  λ$$=1/3$$     Forλ$$=1/3$$     , then M lies outside the segment OB  and hence the required value of λ$$=3$$     .

The line x+y=a meets the axis of x and y at A and B respectively a triangle AMN is inscribed in the triangle OAB, O being the origin, with right angle at N,M and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to

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