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 a tangent to the curve $y=3-x^2$, then the minimum area of triangle $OPQ$, is

• A. 2
• B. 4
• C. 8
• D. 9

Answer 1

Answer: (B)

4

Let $y=f\left(x\right)=3-x^2$

Differentiating with respect to $x$, we get

$f^{\prime }\left(x\right)=-2x$

Consider the equation of tangent at $(x_1,y_1)$

$y-y_1=-2x_1(x-x_1)$

$y-(3-{x_1}^2)=-2x_1(x-x_1)$

$y=-\left(2x_1\right)x+(3+{x_1}^2)$

$y$ intercept $=3+{x_1}^2$

$x$ intercept $=\frac{3+{x_1}^2}{2x_1}$

Since both $x$ and $y$ intercepts are positive, we get $x_1>0$

Area $=\frac{1}{2}.\frac{(3+{x_1}^2{)}^2}{2x_1}$

Let $g\left(x\right)=\frac{(3+{x_1}^2{)}^2}{2x_1}=2\times$ Area

To minimise let us differentiate $g\left(x\right)$

$g^{\prime }\left(x\right)=\frac{-9}{2x^2}+3+\frac{3x^2}{2}$

$g^{\prime }\left(x\right)=0$ $\Rightarrow \frac{-9}{2x^2}+3+\frac{3x^2}{2}=0$

$\Rightarrow 3x^4+6x^2-9=0$

$\Rightarrow (x^2-1)(x^2+3)=0$

$\Rightarrow x=1$ or $x=-1$

Since $x>0$, $x=1$

Now let us check if it is minima or not,

$g^{\prime \prime }\left(x\right)=\frac{9}{x^3}+3x$

$g"\left(1\right)>0$ . Hence $x=1$ is a point of minima.

$2\times Area=g\left(1\right)$

$\Rightarrow Area=\frac{1}{2}.(\frac{9}{2}+3+\frac{1}{2})$

$\Rightarrow Area=4$