Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The equation of the chord joining two points \mathrm{\left(x_1, y_1\right)}  and

\mathrm{\left(x_2, y_2\right)}  on the rectangular hyperbola \mathrm{x y=c^2} is

Option: 1

\mathrm{\frac{x}{x_1+x_2}+\frac{y}{y_1+y_2}=1}


Option: 2

\mathrm{\frac{x}{x_1-x_2}+\frac{y}{y_1-y_2}=1}


Option: 3

\mathrm{\frac{x}{y_1+y_2}+\frac{y}{x_1+x_2}=1}


Option: 4

\mathrm{\frac{x}{y_1-y_2}+\frac{y}{x_1-x_2}=1}


Answers (1)

best_answer

The midpoint of the chord is \mathrm{\left.\left(x_1+x_2\right) / 2,\left(y_1+y_2\right) / 2\right) \text {. }}

The equation of the chord in terms of mid point is 

               \mathrm{T=S_1}

\mathrm{\begin{aligned} & x\left(\frac{y_1+y_2}{2}\right)+y\left(\frac{x_1+x_2}{2}\right)-2 c^2=\left(\frac{x_1+x_2}{2}\right)\left(\frac{y_1+y_2}{2}\right)-2 c^2 \\ & \text { or } x\left(y_1+y_2\right)+y\left(x_1+x_2\right)=\left(x_1+x_2\right)\left(y_1+y_2\right) \\ & \text { or } \frac{x}{x_1+x_2}+\frac{y}{y_1+y_2}=1 \end{aligned}}

Posted by

manish painkra

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 Mains 2026 Registration LIVE: NTA to begin application process soon at jeemain.nta.nic.in; exam date, fee
anam-khan

JEE Mains 2026 Registration LIVE: NTA JEE session 1 application, exam dates soon; city choice available
anam-khan

JEE Main 2026 exam city option not removed: NTA dismisses viral claim

Latest Question