Question

If the length of subnormal is four times the length of subtangent at a point $(3,4)$ on the curve $y=f\left(x\right)$. The tangent at $(3,4)$ to $y=f\left(x\right)$ meets the coordinate axes at $P$ and $Q$ and the maximum area of triangle $OPQ$ (where $O$ is origin) is $4b^2$, then $b=$

• A. $\pm 2$
• B. $\pm \frac{5}{2}$
• C. $\pm \frac{3}{2}$
• D. $\pm \frac{1}{2}$

Answer 1

The correct option is B $\pm \frac{5}{2}$

Length of subtangent at $(3,4)$ $=\left|\frac{y}{m}\right|=\frac{4}{|f^{{}^{\prime }}\left(3\right)|}$
Length of subnormal $=|4m|=|4f^{{}^{\prime }}\left(3\right)|$

Given : $|4m|=4\left|\frac{4}{m}\right|$
$\Rightarrow |m{|}^2=m^2=4\Rightarrow m=\pm 2$


Equation of tangent for $m=2$ at $(3,4)$
$y-4=2(x-3)$
$\Rightarrow 2x-y=2\Rightarrow \frac{x}{1}+\frac{y}{(-2)}=1$
$Ar\left(\Delta OPQ\right)=\frac{1}{2}\times 1\times 2=1$
$\therefore 4b^2=1\Rightarrow b=\pm \frac{1}{2}x$
Equation of tangent for $m=-2$ at $(3,4)$
$(y-4)=-2(x-3)$
$\Rightarrow 2x+y=10\Rightarrow \frac{x}{5}+\frac{y}{10}=1$
$Ar\left(\Delta OPQ\right)=\frac{1}{2}\times 5\times 10=25$
$4b^2=25\Rightarrow b=\pm \frac{5}{2}$
$\therefore b=\pm \frac{5}{2}$