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 $O P Q$, is

• A. 2
• B. 4
• C. 8
• D. 9

Answer 1

Answer: (B)

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