Question

If the area enclosed between $y=m{x}^2$ and $x=n{y}^2$ is $\frac{1}{3}$ sq. units, then $m,n$ can be roots of (where $m,n$ are non zero real numbers)

• A. $2x^2+5x+1=0$
• B. $2x^2+3x-2=0$
• C. $2x^2+3x+2=0$
• D. $2x^2+5x-1=0$

Answer 1

Answer: (A)

$2x^2+5x+1=0$

$y=m{x}^2$

$y^2=m^2{x}^4$

$x=n{m}^2{x}^{1/3}$

$x={\left(\frac{1}{n{m}^2}\right)}^{1/3}$

$y=m\frac{1}{(n{m}^2{)}^{2/3}}$

$y=m\frac{1}{n^{2/3}{m}^{4/3}}$

$y=\frac{1}{(n^{2}m{)}^{1/3}}$

$(1/n{m}^2{)}^{1/3}$

Area$={\int }_0\sqrt{\frac{x}{n}}-m{x}^2$

$={\left[\right]}^{(1/n{m}^2{)}^{1/3}}$

$=\frac{2}{3\sqrt{n}}{\left(\frac{1}{n{m}^2}\right)}^{\frac{1}{3}.\frac{3}{2}}-\frac{m}{3}{\left(\frac{1}{n{m}^2}\right)}^{3\times \frac{1}{3}}$

$=\frac{2}{3\sqrt{n}}\frac{1}{\sqrt{n}m}-\frac{m}{3}\frac{1}{n{m}^2}$

$=\frac{2}{3nm}-\frac{1}{3nm}$

$\frac{1}{3}=\frac{1}{3nm}$

$nm=1$

Product of roots $=1$