Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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.

Posted by

qnaprep

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Latest Question