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The equation of the normal to the curve $\mathrm{x^2=4 y}$ which passes through the point $\mathrm{(1,2)}$ is

Option: 1
$\mathrm{2 x+y=3}$

Option: 2
$\mathrm{x+4 y=3}$

Option: 3
$\mathrm{x+y=3}$

Option: 4
$\mathrm{x+2 y=4}$

Answers (1)

best_answer

$\mathrm{x^2=4 y}$ .....(1)

$\mathrm{ \frac{d y}{d x}=\frac{x}{2} }$

$\mathrm{ {\left[\frac{d y}{d x}\right]_{\left(2 t, t^2\right)}=t} }$

$\therefore \quad$ The slope of the tangent at $\mathrm{P\left(2 t, t^2\right)=t}$

$\mathrm{\therefore}$ The slope of the normal at $\mathrm{P=-\frac{1}{t}}$
Equation of the normal at $\mathrm{P\: is \: y-t^2=-\frac{1}{t}(x-2 t)}$

$\mathrm{ x+y t=2 t+t^3 }$

Since the normal passes through $\mathrm{ (1,2) }$

$\mathrm{ 1+2 t=2 t+t^3 \quad \therefore t^3=1 \text { or } t=1 }$

$\mathrm{ \therefore }$ Required equation of the normal is $\mathrm{ x+y=3 }$

Hence option 3 is correct.

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