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
Verified512.1k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Trending doubts
Which are the Top 10 Largest Countries of the World?
What are the major means of transport Explain each class 12 social science CBSE
Draw a labelled sketch of the human eye class 12 physics CBSE
Why cannot DNA pass through cell membranes class 12 biology CBSE
Differentiate between insitu conservation and exsitu class 12 biology CBSE
Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE