Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The locus of mid-points of the normal chords of the parbola is  <

The locus of mid-points of the normal chords of the parbola \mathrm{y^2=4 a x} is
 

Option: 1

\mathrm{\frac{y^2}{2 a}+\frac{4 a^3}{y^2}=x-2 a}
 


Option: 2

\mathrm{\frac{x^2}{2 a}+\frac{4 a^2}{x^2}=y-2 a}
 


Option: 3

\mathrm{x^3+y^2=x-2 a}
 


Option: 4

\text{ None of these}


Answers (1)

best_answer

Let the equation of the normal be

\mathrm{ y=m x-2 a m-a m^3 }         ........(1)
If its mid-point be \mathrm{(h, k)}, then it is same as

\mathrm{ \mathrm{k}(y-k)=2 a(x-h) }        ...........(2)

\mathrm{ or \: \: y=\frac{2 a}{k} x+\left(k-\frac{2 a h}{k}\right) }

Comparing (1) and (2) we get

\mathrm{ m=\frac{2 a}{k} \text { and } k-\frac{2 a h}{k}=-2 a m-a m^3 }

Eliminating the variable \mathrm{ m } between the above two relations we get

\mathrm{ k-\frac{2 a h}{k}=-2 a\left(\frac{2 a}{k}\right)-a\left(\frac{8 a^3}{k^3}\right) }

\mathrm{ \text { or } k+\frac{8 a^4}{k^3}=\frac{2 a}{k}(h-2 a) \text { or } \frac{k^2}{2 a}+\frac{4 a^3}{k^2}=h-2 a }

Hence the locus of the mid-point \mathrm{ (h, k) } is

\mathrm{ \frac{y^2}{2 a}+\frac{4 a^3}{y^2}=x-2 a }
 

Posted by

Ritika Jonwal

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 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question