Question

If the tangent to the curve $y=1-x^2$ at $x=\alpha$, where $0<\alpha <1$, meets the axes at P and Q. Also $\alpha$ varies, the minimum value of the area of the triangle $OPQ$ is $k$ times the area bounded by the axes and the part of the curve for which $0<x<1$, then $k$ is equal to

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{75}{16}$
• C. $\frac{25}{18}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

$A_1=$ area under the curve and axes
$=\underset{0}{\overset{1}{\int }}\left(1-x^2\right)dx={\left[x-\frac{x^3}{3}\right]}_0^1=\frac{2}{3}$
Now
$\text{ Now }y=1-x^2\therefore y^{\prime }=-2x$
$\therefore y^{\prime }\left(\alpha ,1-{\alpha }^2\right)=-2\alpha$
Equation of tangent to the curve $y=1-x^2$
$\left(y-\left(1-{\alpha }^2\right)\right)=-2\alpha (x-\alpha )$
$2\alpha x+y={\alpha }^2+1$
$\therefore P\equiv \left(\frac{{\alpha }^2+1}{2\alpha },0\right);Q\equiv \left(0,{\alpha }^2+1\right)$
$A=\text{ Area of triangle }POQ=\frac{1}{2}(OP)(OQ)=\frac{1}{4}\frac{{\left({\alpha }^2+1\right)}^2}{\alpha }$
$\therefore A^{\prime }=\frac{1}{4}\frac{\alpha \cdot 2\left({\alpha }^2+1\right)2\alpha -{\left({\alpha }^2+1\right)}^2}{{\alpha }^2}=\frac{1}{4}\frac{3{\alpha }^4+2{\alpha }^2-1}{{\alpha }^2}$
For maximum $/$ minimum, $A^{\prime }=0\Rightarrow 3{\alpha }^4+2{\alpha }^2-1=0\Rightarrow {\alpha }^2=-1,\frac{1}{3}\Rightarrow \alpha =\pm \frac{1}{\sqrt{3}}$
$\therefore 0<\alpha <1\Rightarrow \alpha =\frac{1}{\sqrt{3}},A^{\prime \prime }>0$ hence $A$ is minimum
$A=\frac{1}{4}\frac{{\left(\frac{1}{3}+1\right)}^2}{\frac{1}{\sqrt{3}}}=\frac{4}{3\sqrt{3}}$....(ii)
since $A=kA$ (given)
From (i) and (ii) $\frac{4}{3\sqrt{3}}=k\cdot \frac{2}{3}\Rightarrow k=\frac{2}{\sqrt{3}}$