Question

If the length of subnormal is equal to length of subtangent at any point $(3,4)$ on the curve $y=f\left(x\right)$ and the tangent at $(3,4)$ to $y=f\left(x\right)$ meets the coordinate axes at $A$ and $B$, then maximum area of the $△OAB$ where $O$ is origin, is

• A. $\frac{45}{2}$
• B. $\frac{49}{2}$
• C. $\frac{25}{2}$
• D. $\frac{81}{2}$

Answer 1

The correct option is B $\frac{49}{2}$

$ST=SN$
Hence
$\frac{y}{f^{\prime }\left(x\right)}=y.f^{\prime }\left(x\right)$
Hence
$f^{\prime }(x{)}_{x=3}^2=1$
Or
$f^{\prime }\left(3\right)=\pm 1$
Considering $f^{\prime }\left(3\right)=1$ we get the equation of tangent as
$y-4=(x-3).1$
Or
$y=x+1$ ...(i)
Hence $A=(1,0)$ and $B=(-1,0)$
Thus Area of triangle AOB
$=\frac{1}{2}$
Considering $f^{\prime }\left(3\right)=-1$ we get
$y-4=-1(x-3)$
Or
$y-4=-x+3$
Or
$x+y=7$
Hence
$A=(7,0)$ and $B=(0,7)$.
Thus
Area of AOB
$=\frac{7\times 7}{2}$
$=\frac{49}{2}$
Hence maximum area is $\frac{49}{2}$ sq.unit.