Question

If the length of sub-normal is equal to the length of sub-tangent at 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 the maximum area of the triangle $OAB$, where $O$ represents origin, is

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

Answer 1

Answer: (B)

$\frac{49}{2}$

Given, length of sub-tangent = length of sub-normal
$\left|\frac{y}{m}\right|=|ym|$, where $m$ is the slope at $(3,4)$
$\Rightarrow m=\pm 1$
Equation of tangent at $(3,4)$ and having a slope $m=1$ is
$y-4=1(x-3)$
$y-x=1$
So, the coordinates of $A$ and $B$ are $(-1,0)$ and $(0,1)$ respectively.
So, area of $△OAB=\frac{1}{2}\left(1\right)\left(1\right)=\frac{1}{2}$ sq.units.
Equation of tangent at $(3,4)$ and having a slope $m=-1$ is
$y-4=-1(x-3)$
$y+x=7$
So, coordinates of $A$ and $B$ are $(7,0)$ and $(0,7)$ respectively.
So, area of $△OAB=\frac{1}{2}\left(7\right)\left(7\right)=\frac{49}{2}$ sq.units,
Maximum area of $△OAB=\frac{49}{2}$ sq.units.