Question

lf normal is drawn at a point P (x, y) of a curve meets the X-axis and the Y- axis in point A and B respectively such that $\frac{1}{OA}+\frac{1}{OB}=1$, (where '0' is the origin). The equation of the curve is

• A. $(x-1{)}^2=$ cy
• B. $(x-1{)}^2=$ cx
• C. $(x-1{)}^2+(y-1{)}^2=c$
• D. $x^2+y^2=cx$

Answer 1

Answer: (C)

$(x-1{)}^2+(y-1{)}^2=c$

Normal at a point $p(x,y)$ of a curve meets the x-axis and the y-axis in points A and B receptively S.t. $\frac{1}{OA}+\frac{1}{OB}=\perp$

Normal equation of curve is is

$\frac{dy}{dx}(Y-y)+(X-x)=0$

$\because$ given curve meets x-axis at a point A and Y-axis meets a point B.

$\therefore$ at a point A - y = 0$

So, equation is

$\frac{dy}{dx}(0-y)+(X-x)=0$

$\Rightarrow X=x+y\frac{dy}{dx}$

So, point is $[x+y\frac{dy}{dx},0)$

and similarly, at a point B. $x=0$

So, $\frac{dy}{dx}(Y-y)+(0-x)=0$

$\Rightarrow Y-y=\frac{x}{dy/dx}$

$\Rightarrow Y=y+\frac{x}{dy/dx}$

So, point is $(0,y+\frac{x}{dy/dx})$

Now, find $OA$ and $OB$

$OA=x+y\frac{dy}{dx}$ and $OB=y+\frac{x}{dy/dx}$

and $\frac{1}{OA}+\frac{1}{OB}=\perp \Rightarrow \frac{1}{x+y\frac{dy}{dx}}+\frac{1}{y+\frac{x}{dy/dx}}=\perp$

$\Rightarrow \frac{dx}{xdx+ydy}+\frac{dy}{ydy+xdx}=\perp$

$\Rightarrow dx+dy=xdx+ydy$

$\Rightarrow (x-1)dx+(y-1)dy=0$

Integrating both sides-

$\int (x-1)dx+\int (y-1)dy=c$

$\frac{(x-1{)}^2}{2}+\frac{(y-1{)}^2}{2}=c$

$(x-1{)}^2+(y-1{)}^2=c$

So, option (c) is correct.