Question

The slope of normal at any point $(x,y)$ of a curve $y=f\left(x\right)$, is given by $\frac{-2xy}{x^2+y^2+1}$ and curve passes through $(1,0)$.Then, which of the following point(s) can lie on the curve $y=f\left(x\right)?$

• A. $(3,2\sqrt{2})$
• B. $(5,3\sqrt{2})$
• C. $(\sqrt{8},\sqrt{7})$
• D. $(6,\sqrt{29})$

Answer 1

Answer: (A), (C)

The correct options are
A $(3,2\sqrt{2})$
C $(\sqrt{8},\sqrt{7})$

Slope of normal $=-\frac{1}{\text{slope of tangent}}=-\frac{dx}{dy}$
$-\frac{dx}{dy}=\frac{-2xy}{x^2+y^2+1}\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}2xy\frac{dy}{dx}=x^2+y^2+1$
Now take
$y^2=t\Rightarrow 2y\frac{dy}{dx}=\frac{dt}{dx}\Rightarrow x\frac{dt}{dx}=x^2+t+1\Rightarrow \frac{dt}{dx}+\left(\frac{-t}{x}\right)=\frac{x^2+1}{x}I.F=e^{\int \frac{-1}{x}dx}=\frac{1}{x}t.\frac{1}{x}=\int \frac{1}{x}\left(\frac{x^2+1}{x}\right)dx+c\Rightarrow \frac{y^2}{x}=x-\frac{1}{x}+C$
The curve passes through $(1,0)\Rightarrow C=0$
So, the curve is $x^2-y^2=1$
$(3,2\sqrt{2})\text{and}(\sqrt{8},\sqrt{7})$ satisfies the equation.