Question

Consider the curve $f(x,y)=0$ which satisfies the differential equation $\frac{dy}{dx}+\frac{1}{x-y^2+4}=0$ such that $y\left(1\right)=-1$. If $f(x,y)$ represents a conic, then the length of its latus rectum is

Answer 1

Given:$\frac{dy}{dx}+\frac{1}{x-y^2+4}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{1}{x-y^2+4}$

$\Rightarrow \frac{dx}{dy}=-(x-y^2+4)$
$\Rightarrow \frac{dx}{dy}+x=y^2-4$

Integrate using Integrating factor method
$\Rightarrow x{e}^y=\int e^y(y^2-4)dy$
$\Rightarrow x{e}^y=e^y(y^2-4)-e^y\left(2y\right)+e^y\cdot 2+c$
$\Rightarrow x=(y^2-4-2y+2)+c{e}^{-y}$
$\Rightarrow x=(y-1{)}^2-3+c{e}^{-y}$
Curve passes through $(1,-1).$
$\Rightarrow 1=4-3+c{e}^1$
$\Rightarrow c=0$
$\therefore (y-1{)}^2=(x+3)$