Question

Consider a curve passing through $(1,1)$ such that perpendicular distance of normal drawn at any point $P$ from origin is equal to ordinate of the point $P.$ Then which of the following statement(s) is/are correct?

• A. The differential equation for the curve is $\frac{dy}{dx}=\frac{x^2-y^2}{2xy}.$
• B. The differential equation for the curve is $\frac{dy}{dx}=\frac{y^2-x^2}{2xy}.$
• C. The equation of tangent at $(2,0)$ is $x=2.$
• D. The curve passes through origin.

Answer 1

Answer: (B), (C), (D)

The curve passes through origin.

Let $(x_0,y_0)$ be the point $P.$
Figure:


Equation of normal at point $P$ is
$y-y_0=-\frac{1}{m}(x-x_0)$
$\Rightarrow my-m{y}_0=-x+x_0$
$\Rightarrow x+my-m{y}_0-x_0=0$
given, perpendicular distance of normal drawn at any point $P$ from origin is equal to ordinate of the point $P.$
$\therefore OQ=\frac{0+0-m{y}_0-x_0}{\sqrt{(1+m^2)}}=y_0$
$\Rightarrow m^2{y}_0^2+x_0^2+2m{y}_0{x}_0=y_0^2+y_0^2{m}^2$
$\Rightarrow 2xy\frac{dy}{dx}-y^2=-x^2$
$\Rightarrow 2y\frac{dy}{dx}-\frac{y^2}{x}=-x$
Put $t=y^2\Rightarrow \frac{dt}{dx}=2y\frac{dy}{dx}$
$\therefore \frac{dt}{dx}-\frac{t}{x}=-x$
$I.F.=e^{-\ln x}=\frac{1}{x}$
Thus, general solution is
$t\cdot \frac{1}{x}=-\int x\cdot \frac{1}{x}dx$
$\Rightarrow y^2=-x^2+Cx$
Since, it passes through $(1,1),$ therefore $C=2$
$\therefore y^2=-x^2+2x$
$\Rightarrow x^2+y^2-2x=0$
Hence, curve passes through $(0,0).$


We have, $x^2+y^2-2x=0$
The equation of tangent at $(2,0)$ is
$x\left(2\right)+y\left(0\right)-2\left(\frac{x+2}{2}\right)=0$
$\Rightarrow x=2$