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Q.
Three normals to the parabola $y^{2}=x$ are drawn a point $(c, 0)$, then :
Bihar CECEBihar CECE 2003
Solution:
The slope form of the normal equation to the parabola
$y^{2}=4 a x$ is
$y=m x-2 a m-a m^{3}$
The slope form of the normal to the parabola $y^{2}=4 a x$ is
$y=m x-2 a m-a m^{3}$
Since, the given curve is $y^{2}=x$
Here, $a=\frac{1}{4}$
$\therefore y=m x-\frac{1}{2} m-\frac{1}{4} m^{3}$
If it passes through $(c, 0)$, then
$0=m c-\frac{1}{2} m-\frac{1}{4} m^{3}$
$\Rightarrow m=0$ or $c-\frac{1}{2}-\frac{1}{4} m^{2}=0$
$\Rightarrow m=\pm 2 \sqrt{c-\frac{1}{2}}$
For three normal values of $m$, it should be real.
$\therefore c>\frac{1}{2}$.