Find an equation of a curve passing through the point $\left( 1,1 \right)$ if the perpendicular distance of the origin from the normal at any point $p\left( x,y \right)$ of the curve is equal to the distance of $p$ from the X-axis.
Answer
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Hint: We first take the equation of the normal and find its distance from the origin. We equate it with the distance of $p\left( x,y \right)$ from the X-axis. We solve the differential equation and find the equation which satisfies the point $\left( 1,1 \right)$ as the curve passes through it.
Complete answer:
Let us assume the curve is $y=f\left( x \right)$. The slope of the normal will be $-\dfrac{dx}{dy}$ where $\dfrac{dy}{dx}={{f}^{'}}\left( x \right)$.
We are trying to find the normal at point $p\left( x,y \right)$.
The equation of the normal becomes $Y-y=\left( -\dfrac{dx}{dy} \right)\left( X-x \right)$.
It is given that the perpendicular distance of the origin from the normal at any point $p\left( x,y \right)$ of the curve is equal to the distance of $p$ from the X-axis.
The distance of $p\left( x,y \right)$ from the X-axis is $y$.
We also know the distance formula from a point $\left( m,n \right)$ to the line $Ax+By-c=0$ will be equal to $\left| \dfrac{Am+Bn-c}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$ units.
Therefore, the perpendicular distance of the origin $\left( 0,0 \right)$ from the normal $Y-y=\left( -\dfrac{dx}{dy} \right)\left( X-x \right)$ will be $\left| \dfrac{-y-x\dfrac{dx}{dy}}{\sqrt{1+{{\left( \dfrac{dx}{dy} \right)}^{2}}}} \right|$.
Forming the equation, we get $\left| \dfrac{-y-x\dfrac{dx}{dy}}{\sqrt{1+{{\left( \dfrac{dx}{dy} \right)}^{2}}}} \right|=y$.
We take the square and simplify it.
\[\begin{align}
& {{\left( -y-x\dfrac{dx}{dy} \right)}^{2}}={{y}^{2}}\left( 1+{{\left( \dfrac{dx}{dy} \right)}^{2}} \right) \\
& \Rightarrow {{y}^{2}}+{{x}^{2}}{{\left( \dfrac{dx}{dy} \right)}^{2}}+2xy\dfrac{dx}{dy}={{y}^{2}}+{{y}^{2}}{{\left( \dfrac{dx}{dy} \right)}^{2}} \\
& \Rightarrow {{\left( \dfrac{dx}{dy} \right)}^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)+2xy\dfrac{dx}{dy}=0 \\
& \Rightarrow \dfrac{dx}{dy}\left[ \dfrac{dx}{dy}\left( {{x}^{2}}-{{y}^{2}} \right)+2xy \right]=0 \\
\end{align}\]
Solving it we get either \[\dfrac{dx}{dy}=0\] or \[\dfrac{dy}{dx}=\dfrac{{{y}^{2}}-{{x}^{2}}}{2xy}\].
The solution of first differential form gives $x=c$ but that doesn’t satisfy the given conditions.
Now we take \[\dfrac{dy}{dx}=\dfrac{{{y}^{2}}-{{x}^{2}}}{2xy}=\dfrac{1}{2}\left( \dfrac{y}{x}-\dfrac{x}{y} \right)\]. We assume \[\dfrac{y}{x}=v\].
The differential of \[y=vx\] gives \[\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\]. Replacing we get
\[\begin{align}
& \dfrac{dy}{dx}=\dfrac{1}{2}\left( \dfrac{y}{x}-\dfrac{x}{y} \right) \\
& \Rightarrow v+x\dfrac{dv}{dx}=\dfrac{1}{2}\left( v-\dfrac{1}{v} \right) \\
& \Rightarrow 2x\dfrac{dv}{dx}=\left( -v-\dfrac{1}{v} \right) \\
& \Rightarrow \dfrac{2vdv}{{{v}^{2}}+1}=-\dfrac{dx}{x} \\
\end{align}\]
We use the differential form of $d\left( {{v}^{2}}+1 \right)=2vdv$.
\[\begin{align}
& \dfrac{2vdv}{{{v}^{2}}+1}=-\dfrac{dx}{x} \\
& \Rightarrow \dfrac{d\left( {{v}^{2}}+1 \right)}{{{v}^{2}}+1}+\dfrac{dx}{x}=0 \\
\end{align}\]
Then we use integration to get \[\int{\dfrac{d\left( {{v}^{2}}+1 \right)}{{{v}^{2}}+1}}+\int{\dfrac{dx}{x}}=k\]. Here $k$ is integral constant.
\[\begin{align}
& \int{\dfrac{d\left( {{v}^{2}}+1 \right)}{{{v}^{2}}+1}}+\int{\dfrac{dx}{x}}=k \\
& \Rightarrow \log \left| {{v}^{2}}+1 \right|+\log \left| x \right|=k \\
\end{align}\]
We put value of \[v=\dfrac{y}{x}\] in the equation to get
\[\begin{align}
& \log \left| {{v}^{2}}+1 \right|+\log \left| x \right|=k \\
& \Rightarrow \log \left| \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right|+\log \left| x \right|=k \\
\end{align}\]
The curve goes through point $\left( 1,1 \right)$. Putting the value we get \[k=\log \left| \dfrac{2}{1} \right|+\log \left| 1 \right|=\log 2\].
So, \[\log \left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right)+\log \left| x \right|=\log 2\]. We multiply 2 both sides to get
\[\begin{align}
& 2\log \left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right)+2\log \left| x \right|=2\log 2 \\
& \Rightarrow \log {{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right)}^{2}}+\log {{x}^{2}}=\log 4 \\
& \Rightarrow \log {{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{x} \right)}^{2}}=\log 4 \\
& \Rightarrow {{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{x} \right)}^{2}}=4 \\
\end{align}\]
The equation becomes \[{{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{x} \right)}^{2}}={{\left( \pm 2 \right)}^{2}}\Rightarrow {{y}^{2}}+{{x}^{2}}=\pm 2x\].
Now putting value of $\left( 1,1 \right)$ in equation \[{{y}^{2}}+{{x}^{2}}=-2x\], we can see it doesn’t satisfy.
So, the equation of the curve is \[{{y}^{2}}+{{x}^{2}}=2x\].
Note:
In the differential equation we solved it by removing the modulus function. Else we have to use both the signs to proceed. We cannot take square as in logarithm the terms ${{\left( \log x \right)}^{2}}$ and $\log {{x}^{2}}$ denote different things.
Complete answer:
Let us assume the curve is $y=f\left( x \right)$. The slope of the normal will be $-\dfrac{dx}{dy}$ where $\dfrac{dy}{dx}={{f}^{'}}\left( x \right)$.
We are trying to find the normal at point $p\left( x,y \right)$.
The equation of the normal becomes $Y-y=\left( -\dfrac{dx}{dy} \right)\left( X-x \right)$.
It is given that the perpendicular distance of the origin from the normal at any point $p\left( x,y \right)$ of the curve is equal to the distance of $p$ from the X-axis.
The distance of $p\left( x,y \right)$ from the X-axis is $y$.
We also know the distance formula from a point $\left( m,n \right)$ to the line $Ax+By-c=0$ will be equal to $\left| \dfrac{Am+Bn-c}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$ units.
Therefore, the perpendicular distance of the origin $\left( 0,0 \right)$ from the normal $Y-y=\left( -\dfrac{dx}{dy} \right)\left( X-x \right)$ will be $\left| \dfrac{-y-x\dfrac{dx}{dy}}{\sqrt{1+{{\left( \dfrac{dx}{dy} \right)}^{2}}}} \right|$.
Forming the equation, we get $\left| \dfrac{-y-x\dfrac{dx}{dy}}{\sqrt{1+{{\left( \dfrac{dx}{dy} \right)}^{2}}}} \right|=y$.
We take the square and simplify it.
\[\begin{align}
& {{\left( -y-x\dfrac{dx}{dy} \right)}^{2}}={{y}^{2}}\left( 1+{{\left( \dfrac{dx}{dy} \right)}^{2}} \right) \\
& \Rightarrow {{y}^{2}}+{{x}^{2}}{{\left( \dfrac{dx}{dy} \right)}^{2}}+2xy\dfrac{dx}{dy}={{y}^{2}}+{{y}^{2}}{{\left( \dfrac{dx}{dy} \right)}^{2}} \\
& \Rightarrow {{\left( \dfrac{dx}{dy} \right)}^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)+2xy\dfrac{dx}{dy}=0 \\
& \Rightarrow \dfrac{dx}{dy}\left[ \dfrac{dx}{dy}\left( {{x}^{2}}-{{y}^{2}} \right)+2xy \right]=0 \\
\end{align}\]
Solving it we get either \[\dfrac{dx}{dy}=0\] or \[\dfrac{dy}{dx}=\dfrac{{{y}^{2}}-{{x}^{2}}}{2xy}\].
The solution of first differential form gives $x=c$ but that doesn’t satisfy the given conditions.
Now we take \[\dfrac{dy}{dx}=\dfrac{{{y}^{2}}-{{x}^{2}}}{2xy}=\dfrac{1}{2}\left( \dfrac{y}{x}-\dfrac{x}{y} \right)\]. We assume \[\dfrac{y}{x}=v\].
The differential of \[y=vx\] gives \[\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\]. Replacing we get
\[\begin{align}
& \dfrac{dy}{dx}=\dfrac{1}{2}\left( \dfrac{y}{x}-\dfrac{x}{y} \right) \\
& \Rightarrow v+x\dfrac{dv}{dx}=\dfrac{1}{2}\left( v-\dfrac{1}{v} \right) \\
& \Rightarrow 2x\dfrac{dv}{dx}=\left( -v-\dfrac{1}{v} \right) \\
& \Rightarrow \dfrac{2vdv}{{{v}^{2}}+1}=-\dfrac{dx}{x} \\
\end{align}\]
We use the differential form of $d\left( {{v}^{2}}+1 \right)=2vdv$.
\[\begin{align}
& \dfrac{2vdv}{{{v}^{2}}+1}=-\dfrac{dx}{x} \\
& \Rightarrow \dfrac{d\left( {{v}^{2}}+1 \right)}{{{v}^{2}}+1}+\dfrac{dx}{x}=0 \\
\end{align}\]
Then we use integration to get \[\int{\dfrac{d\left( {{v}^{2}}+1 \right)}{{{v}^{2}}+1}}+\int{\dfrac{dx}{x}}=k\]. Here $k$ is integral constant.
\[\begin{align}
& \int{\dfrac{d\left( {{v}^{2}}+1 \right)}{{{v}^{2}}+1}}+\int{\dfrac{dx}{x}}=k \\
& \Rightarrow \log \left| {{v}^{2}}+1 \right|+\log \left| x \right|=k \\
\end{align}\]
We put value of \[v=\dfrac{y}{x}\] in the equation to get
\[\begin{align}
& \log \left| {{v}^{2}}+1 \right|+\log \left| x \right|=k \\
& \Rightarrow \log \left| \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right|+\log \left| x \right|=k \\
\end{align}\]
The curve goes through point $\left( 1,1 \right)$. Putting the value we get \[k=\log \left| \dfrac{2}{1} \right|+\log \left| 1 \right|=\log 2\].
So, \[\log \left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right)+\log \left| x \right|=\log 2\]. We multiply 2 both sides to get
\[\begin{align}
& 2\log \left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right)+2\log \left| x \right|=2\log 2 \\
& \Rightarrow \log {{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{{{x}^{2}}} \right)}^{2}}+\log {{x}^{2}}=\log 4 \\
& \Rightarrow \log {{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{x} \right)}^{2}}=\log 4 \\
& \Rightarrow {{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{x} \right)}^{2}}=4 \\
\end{align}\]
The equation becomes \[{{\left( \dfrac{{{y}^{2}}+{{x}^{2}}}{x} \right)}^{2}}={{\left( \pm 2 \right)}^{2}}\Rightarrow {{y}^{2}}+{{x}^{2}}=\pm 2x\].
Now putting value of $\left( 1,1 \right)$ in equation \[{{y}^{2}}+{{x}^{2}}=-2x\], we can see it doesn’t satisfy.
So, the equation of the curve is \[{{y}^{2}}+{{x}^{2}}=2x\].
Note:
In the differential equation we solved it by removing the modulus function. Else we have to use both the signs to proceed. We cannot take square as in logarithm the terms ${{\left( \log x \right)}^{2}}$ and $\log {{x}^{2}}$ denote different things.
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