Question

The area of the figure bounded by a curve, the x-axis, and two ordinates, one of which is constant the other variable, is equal to ratio of the cube of the variable ordinate to the variable abscissa. The curve is given by

• A. $(2y^2-x^2{)}^2=C{x}^2$
• B. $(2y^2-x^2{)}^3=C{x}^2$
• C. $(y^2-x^2{)}^3=C{x}^2$
• D. $(2y-x^2{)}^3=C{x}^2$

Answer 1

Answer: (B)

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

Area of the shaded region ${\int }_a^{h}f\left(x\right)dx=\frac{f^3\left(h\right)}{h}$ (given)
Differentiating both sides w.r.t 'h', we get
$f\left(h\right)=\frac{h3t^2\left(h\right).f^{\prime }\left(h\right)-f^2\left(h\right)}{h^2}\Rightarrow h^{2}f\left(h\right)=3h{f}^2\left(h\right)f^{\prime }\left(h\right)-f^3\left(h\right)$ ...(1)
Replace $f\left(h\right)$ by $y$ and $h$ by $x$
$x^{2}y=3xy{y}^2\frac{dy}{dx}-y^3\Rightarrow \frac{dy}{dx}=\frac{x^{2}y+y^3}{3x{y}^2}$ ...(2)
Equation (2) is a homogeneous differential equation
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$ in (2)

$\Rightarrow v+x\frac{dv}{dx}=\frac{v+v^3}{3v^2}\Rightarrow \frac{3v^{2}dv}{2v^3-v}=-\frac{dx}{x}$

$\Rightarrow \frac{3v}{2v^3-v}=-\frac{dx}{x}\Rightarrow \frac{3}{4}\int \frac{4vdv}{2v^2-1}=-\int \frac{dx}{x}$

$\Rightarrow \frac{3}{4}\log (2v^2-1)=\log \left(\frac{c}{x}\right)\Rightarrow {(2v^2-1)}^{\frac{3}{4}}=\frac{c}{x}$

$\because v=\frac{y}{x}\Rightarrow {(2y^2-x^2)}^3=c{x}^2$