Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Answers (1)

best_answer

Equation of hyperbolas having foci on x-axis and centre at the origin
\frac{x^2}{b^2}+\frac{y^2}{a^2} = 1                 
Now, differentiate w..r.t.  x
\frac{2x}{b^2}+\frac{2y}{a^2}.\frac{dy}{dx}=0\\                                    -(i)
Now, again differentiate w.r.t. x
\frac{2}{b^2}+\frac{2}{a^2}.y^{'}.y^{'}+\frac{2y}{a^2}.y^{''}=0\\ \\ \frac{1}{b^2}=-\frac{1}{a^2}\left ( (y^{'})^2+yy^{''} \right )         -(ii)
Put value from equation (ii) in (i)
Our equation becomes
\frac{2y}{a^2}y^{'}-\frac{2x}{a^2}\left ( (y^{'})^2+yy^{''} \right )=0\\ \\ 2yy^{'}-2(y^{'})^2x+2yy^{''}x=0\\ \\ xyy^{''}-x(y^{'})^2+yy^{'}= 0
Therefore, the required equation is   xyy^{''}-x(y^{'})^2+yy^{'}= 0

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question