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

The correct option is A $x=0$

Slope of the normal $=\frac{y}{x-1}$
$\therefore \frac{dy}{dx}=\frac{1-x}{y}$
$\frac{y^2}{2}=x-\frac{x^2}{2}+C$
Eq (ii) passes through $(0,0)$
Thus $C=0$
$x^2+y^2-2x=0$
Now, tangent to $y^2=4x$
$y=mx+\frac{1}{m}$
If it touches the circle
$x^2+y^2-2x=0$
Then, $\left|\frac{m+\left(1/m\right)}{\sqrt{1+m^2}}\right|=1$
$\Rightarrow 1+m^2=m^2$
$\Rightarrow m\to \infty$
Hence, tangent is $y-$ axs i.e., $x=0$