Question

A curve passing through the point $(1,1)$ has the property that the perpendicular distance of the origin from the normal at any point $P$ of the curve is equal to the distance of $P$ from the x-axis. The equation of the curve represents

• A. $x^2+y^2=2x$
• B. $x^2+y^2=2y$
• C. $x^2+y^2=2$
• D. None of these

Answer 1

The correct option is A $x^2+y^2=2x$

Equation of normal at point $(x,y)$ is

$Y-y=-\frac{dy}{dx}(X-x)$ ...(1)

Distance of perpendicular from the origin to Eq.(1)

$=\frac{|y+\frac{dx}{dy}.x|}{\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}$

also, distance between $P$ and x-axis is $\left|y\right|$

$\therefore =\frac{|y+\frac{dx}{dy}.x|}{\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}=\left|y\right|$

$\Rightarrow y^2+\frac{dx}{dy}.x^2+2xy\frac{dx}{dy}=y^2[1+{\left(\frac{dx}{dy}\right)}^2]$

$\Rightarrow {\left(\frac{dx}{dy}\right)}^2(x^2+y^2)+2xy\frac{dx}{dy}=0$

$\Rightarrow \frac{dx}{dy}[\left(\frac{dx}{dy}\right)(x^2-y^2)+2xy]=0$

$\Rightarrow \frac{dx}{dy}=0\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$

But $\frac{dx}{dy}=0$

$\Rightarrow x=c,$ where c is a constant.

Since, curve passes through $(1,1),$ we get the equation of the cure as $x=1.$

The equation $\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ is a homogeneous equation.

Substitute $y=vx\Rightarrow \frac{dv}{dx}=v+x\frac{dv}{dx}v+x\frac{dv}{dx}=\frac{v^2-1-x^2}{{2x}^{2}v}$

$\Rightarrow x\frac{dv}{dx}=\frac{v^2-1-x^2}{{2x}^{2}v}=-\frac{v^2+1}{2v}$

$\Rightarrow \frac{-2v}{v^2+1}dv=\frac{dx}{x}$

$\Rightarrow c_1-\log (v^2+1)=\log \left|x\right|$

$\Rightarrow \log \left|x\right|(v^2+1)=c_1\Rightarrow \left|x\right|(\frac{y^2}{x^2}+1)=e^{c_1}$

$\Rightarrow x^2+y^2=\pm e^{c_1}x$ or $x^2+y^2=\pm e^{c}x$ is passing through $(1,1).$

$\therefore 1+1=\pm e^c.1\Rightarrow \pm e^c=2.$

Hence, required curve is $x^2+y^2=2x.$