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 $y(1)=-1$. If $f(x,y)$ represents a conic then find the length of its latus rectum.

Answer 1

Answer: 1

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