Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If  are the intercept of a focal chord of the parabola <img alt="\mathrm{ y^2=4 a x

If \mathrm{b, k } are the intercept of a focal chord of the parabola \mathrm{ y^2=4 a x }, then \mathrm{ K} is equal to 

Option: 1

\mathrm{\frac{a b}{b-a}}


Option: 2

\mathrm{\frac{b}{b-a}}


Option: 3

\mathrm{ \frac{a}{b-a}}


Option: 4

\mathrm{\frac{a b}{a-b}}


Answers (1)

best_answer

Let  \mathrm{' t_1{ }^{\prime}}  \mathrm{'t_2{ }^{\prime}}  be the ends of focal chords 

\mathrm{\therefore t_1 t_2=-1 }. If S is the focus and  P, Q are the ends of the focal chord, then 

\mathrm{ S P=\sqrt{\left(a t_1^2-a\right)^2+\left(2 a t_1-0\right)^2}=a\left(t_1^2+1\right)=b \, \, \, \, \, (Given)... (i)}

\mathrm{\therefore S Q=a\left(t_2^2+1\right) \quad-a\left(\frac{1}{t_1^2}+1\right) \quad \text { (Given) } \left[\ t_2=-\frac{1}{t_1} \Rightarrow t_2^2=\frac{1}{t_1^2}\right]}

\mathrm{=\frac{a\left(t_1^2+1\right)}{t_1^2}=k}        \mathrm{\text { ....(ii), }}      \mathrm{\therefore \frac{b}{k}=t_1^2 \quad \text { [Divide (i) by (ii)] }}


Putting in (1), we get \mathrm{a\left(\frac{b}{k}+1\right)=b \Rightarrow \frac{a b}{k}+a=b \quad k=\frac{a b}{b-a}}

Posted by

Ajit Kumar Dubey

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question