Question

A curve $C$ passes through origin and has the property that at each point $(x,y)$ on it the normal line at that point passes through $(1,0)$. The equation of a common tangent to the curve $C$ and the parabola $y^2=4x$ is

• A. $x=0$
• B. $y=0$
• C. $y=x+1$
• D. $x+y+1=0$

Answer 1

Answer: (A)

Slope of the normal $=\frac{y}{x-1}$
$\therefore \frac{dy}{dx}=\frac{1-x}{y}$
$\frac{y^2}{2}=x-\frac{x^2}{2}+C$....(2)

(2) passes through $(0,0)$ hence $C=0$
$x^2+y^2-2x=0$
now tangent to $y^2=4x$
$y=mx+\frac{1}{m}$....(3)

if it touches the circle
$x^2+y^2-2x=0$
then $\left|\frac{m+(1/m)}{\sqrt{1+m^2}}\right|=1\Rightarrow 1+m^2=m^2\Rightarrow m\to \infty$
hence tangent is $y$ axis i.e. $x=0$