Question

A normal at any point $(x,y)$ to the curve $y=f\left(x\right)$ cuts a triangle of unit area with the co-ordinate axes, then the differential equation of the curve is:

• A. $y^2-x^2{\left(\frac{dy}{dx}\right)}^2=4\frac{dy}{dx}$
• B. $x^2-y^2{\left(\frac{dy}{dx}\right)}^2=\frac{dy}{dx}$
• C. $x+y\frac{dy}{dx}=y$
• D. $y^2{\left(\frac{dy}{dx}\right)}^2+2(xy-1)\frac{dy}{dx}+x^2=0$

Answer 1

The correct option is D $y^2{\left(\frac{dy}{dx}\right)}^2+2(xy-1)\frac{dy}{dx}+x^2=0$

Equation of normal at point $p$ is $Y-y=-\frac{dx}{dy}(X-x)$
So, the intercepts are $A(y\frac{dy}{dx}+x,0)$ and $B(0,y+x\frac{dx}{dy})$:
Area of $\Delta OAB\text{is}1\text{unit}$
$\Rightarrow \frac{1}{2}\left|(y\frac{dy}{dx}+x)(x\frac{dx}{dy}+y)\right|=1$
$\Rightarrow (y\frac{dy}{dx}+x)(y\frac{dy}{dx}+x)=\pm 2\frac{dy}{dx}$
$\Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2+2(xy\pm 1)\frac{dy}{dx}+x^2=0$
Based on given options, only 4th option is correct.