Question

$P$ is a point on the positive $x-$axis, $Q$ is a point on the positive $y-$axis and $O$ is the origin. If the line passing through $P$ and $Q$ is tangent to the curve $y=3-x^2,$ then the minimum area of triangle $OPQ$ is

• A. $16$ sq. units
• B. $4$ sq. units
• C. $1$ sq. unit
• D. $2$ sq. units

Answer 1

The correct option is B $4$ sq. units

Let $T(a,3-a^2)$ be the point of contact on curve $y=3-x^2$
$\frac{dy}{dx}=-2a$
Equation of tangent at $T$ is
$y-(3-a^2)=-2a(x-a)$
$\Rightarrow 2ax+y=2a^2+3-a^2$
$\Rightarrow 2ax+y=a^2+3$


For $P,$ $y=0,x=\frac{a^2+3}{2a}$
For $Q,$ $x=0,y=a^2+3$
Area of $△OPQ=\frac{1}{2}\cdot \frac{{(a^2+3)}^2}{2a}$

Let $f\left(a\right)=\frac{{(a^2+3)}^2}{4a}$
$f^{\prime }\left(a\right)=\frac{1}{4}\left(\frac{2\cdot 2a\cdot a(a^2+3)-(a^2+3{)}^2}{a^2}\right)=0$
$\Rightarrow (a^2+3)(4a^2-a^2-3)=0$
$\Rightarrow a^2=1$
$\Rightarrow a=1$ or $-1$
But $a=-1$ is not possible.
By first derivative test,
$f\left(a\right)$ has local minimum at $a=1$
$\therefore A_{min}=f\left(1\right)=\frac{16}{4}=4$ sq. units