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Q.
Distance of the normal to the curve $x^{2}=4 y$ which passes through the point $(1,2)$ is
Conic Sections
Solution:
Equation of normal to $x^{2}=4 a y$ in standard form is
$ x=m y-2 a m-a m^{3}$
$\therefore$ Equation of normal to $x^{2}=4 y$ is
$x=m y-2 m-m^{3}$
It passes through the point $(1,2)$
$\therefore 1=2 m-2 m-m^{3}$
$\therefore m=-1$
$\therefore $ Equation of normal is $y-2=-(x-1)$
or $x+y-3=0$
Its distance from origin is $\frac{3}{\sqrt{2}}$.