A point M on a curve (c) projects onto the y -axis at Q…
A point $M$ on a curve $(c)$ projects onto the $y$-axis at $Q$, and the normal at $M$ meets the $x$-axis at $N$. Find the curves $(c)$ for which $OM\perp QN$, and the member through $(2,0)$.
Explanation of Method
Express $N$ from the normal line's equation, form the slopes of $OM$ and $QN$, and impose the perpendicularity condition (product of slopes $=-1$) to get a homogeneous ODE.
Step 1: Locate $Q$ and $N$
$Q=(0,y)$. The normal at $M=(x,y)$ has slope $-1/y'$, so it meets the $x$-axis at $N=(x+yy',0)$.
Step 2: Impose perpendicularity
Slope of $OM$ is $y/x$; slope of $QN$ is $\dfrac{-y}{x+yy'}$. Setting their product to $-1$:
\[\frac{y}{x}\cdot\frac{-y}{x+yy'}=-1 \quad\Longrightarrow\quad y^2=x^2+xyy' \quad\Longrightarrow\quad y'=\frac{y^2-x^2}{xy}.\]
Step 3: Solve with $y=vx$
\[x\frac{dv}{dx}=\frac{v^2-1}{v}-v=-\frac1v \quad\Longrightarrow\quad v\,dv=-\frac{dx}{x} \quad\Longrightarrow\quad v^2=C-2\ln|x|.\]
Back-substituting $v=y/x$:
\[\boxed{\,y^2=x^2(C-2\ln|x|).}\]
Step 4: Apply the point condition
At $(2,0)$: $0=4(C-2\ln2)\Rightarrow C=2\ln2$, giving
\[\boxed{\,y^2=2x^2\ln\frac{2}{x},\qquad 0 Final answer: General family: $y^2=x^2(C-2\ln|x|)$. Through $(2,0)$: \[\boxed{\,y^2=2x^2\ln\dfrac2x.}\]
Result