Question

P is a point on 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 the triangle $OPQ$, is

• A. 2
• B. 4
• C. 8
• D. 9

Answer 1

Answer: (B)

$y=3-x^2$
${\frac{dy}{dx}|}_T=-2x=-2a$
equation of tangent at $T$ is
$y-\left(3-a^2\right)=-2a(x-a)$
$2ax+y=2a^2+3-a^2=a^2+3$
$y=0,x=\frac{a^2+3}{2a};x=0,y=a^2+3$
$\text{ Area of }OPQ=\frac{1}{2}\cdot \frac{{\left(a^2+3\right)}^2}{2a};\text{ Let }f(a)=\frac{{\left(a^2+3\right)}^2}{4a}$
$f^{\prime }(a)=\frac{1}{4}\left[\frac{2a^2\cdot 2\left(a^2+3\right)-{\left(a^2+3\right)}^2}{a^2}\right]=0$
$\left(a^2+3\right)\left(4a^2-a^2-3\right)=0$
$a^2=1\Rightarrow a=1\text{ or }-1$
$\therefore A_{\min }=\frac{16}{4}=4\text{ sq. units . }$