Question

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

Answer 1

$ \frac{ dy }{ dx }=\frac{ l }{ y ^2- x -4} \Rightarrow \frac{ dx }{ dy }+ x = y ^2-4 \Rightarrow x \cdot e ^{ y }=\int e ^{ y }\left( y ^2-4\right) dy + c$
$x \cdot e ^{ y }= e ^{ y }\left( y ^2-4\right)- e ^{ y }(2 y )+ e ^{ y } \cdot 2+ c $
$x =\left( y ^2-4-2 y +2\right)+ c \cdot e ^{- y } $
$x =( y -1)^2-3+ c \cdot e ^{- y } $
$\text { passes }(1,-1) \Rightarrow 1=4-3+ c \cdot e ^{- y }$
$c =0 $
$\therefore(y-1)^2=(x+3) $