Question

The line x + y = a meets the x-axis of A and y-axis at B. 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 area of DAMN is $\frac { 3 } { 8 }$ of the area of DOAB, then $\frac { \mathrm { AN } } { \mathrm { BN } }$ is equal to-

• A. 3
• B. $\frac { 1 } { 3 }$
• C. 3 or $\frac { 1 } { 3 }$
• D. $\frac { 2 } { 3 }$

Answer 1

Answer: (A)

3

Answer 2

Let = l, Then N ŗ $\left( \frac { \mathrm { a } } { 1 + \lambda } , \frac { \lambda \mathrm { a } } { 1 + \lambda } \right)$

Slope of MN is 1, so, equation of MN is

y –$\frac { \lambda \mathrm { a } } { 1 + \lambda }$ = x – … (i)

Hence, M is

Area of D AMN = $\frac { 1 } { 2 }$AN × MN

= $\frac { 1 } { 2 }$=

Area of D OAB = $\frac { 1 } { 2 }$a² . According to given condition

$\frac { 1 } { 2 }$ a² Ž 3x² – 10l + 3 = 0

Ž l = 3, $\frac { 1 } { 3 }$.

For l = $\frac { 1 } { 3 }$, M lies outside the segment OB, hence l = 3.