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

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