Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The normal chord to a parabola $\mathrm{y^2=4 a x}$ at the point whose ordinate is equal to abscissa subtends an angle $\mathrm{\theta}$ at the focus, then $\mathrm{\frac{\theta}{2}=}$

Option: 1
$\frac{\pi}{6}$

Option: 2
$\frac{\pi}{4}$

Option: 3
$\frac{\pi}{3}$

Option: 4
$\frac{\pi}{2}$

Answers (1)

best_answer

Let the normal at $\mathrm{P\left(a t_1^2, 2 a t_1\right)}$ meets the parabola again at $\mathrm{Q\left(a t_2^2, 2 a t_2\right)}$ .

$\mathrm{\therefore \quad P Q}$ is a normal chord and

$\mathrm{ t_2=-t_1-\frac{2}{t_1} }$ ......(i)

$\mathrm{ Given \: \: 2 a t_1=a t_1^2 \Rightarrow t_1=2 }$

$\mathrm{ \therefore }$ From (i), $\mathrm{ t_2=-3 }$

Thus, $\mathrm{ P \equiv(4 a, 4 a) and Q \equiv(9 a,-6 a) }$

$\mathrm{ \therefore \quad \: Slope\: of \: S P=\frac{4 a-0}{4 a-a}=\frac{4}{3} }$

$\mathrm{and\: slope\: of \: S Q=\frac{-6 a-0}{9 a-a}=\frac{-3}{4}}$

$\mathrm{Since \: slope \: of\: S P \times\: slope of \: S Q=-1}$

$\mathrm{\therefore \angle P S Q=\frac{\pi}{2} i.e., P Q subtends\: a \, right \: \: angle \: at\: the \: focus\: S}$

Hence option 2 is correct.

Posted by

himanshu.meshram

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

Trending Articles/News

anam-khan

NTA Exams 2025 Live: NEET UG date, registration form soon at neet.nta.nic.in; UGC NET revised schedule out

anam-khan

Is it difficult to get into NIT Durgapur? JEE Main cut-offs for BTech programmes

anam-khan

JEE Main Deleted Syllabus 2025: Topics to skip this year; why no optional questions?

Latest Question