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 the middle points of chords of the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7

The locus of the middle points of chords of the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} which pass through a fixed point \mathrm{(\alpha, \beta)} is a hyperbola whose centre is

Option: 1

\left(\frac{\alpha}{3}, \frac{\beta}{3}\right)


Option: 2

(\alpha, \beta)


Option: 3

\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)


Option: 4

\left(\frac{\beta}{2}, \frac{\alpha}{2}\right)


Answers (1)

best_answer

Equation of the chord of the hyperbola whose midpoint is (h, k) given by \mathrm{T=S_1} i.e.,

\mathrm{ \frac{h x}{a^2}-\frac{k y}{b^2}=\frac{h^2}{a^2}-\frac{k^2}{b^2} \text {. } }
It passes through the point \mathrm{(\alpha, \beta)}

\mathrm{ \begin{aligned} & \therefore \frac{h \alpha}{a^2}-\frac{k \beta}{b^2}=\frac{h^2}{a^2}-\frac{k^2}{b^2} \\\\ & \therefore \text { Locus of }(h, k) \text { is } \frac{x^2-\alpha x}{a^2}-\frac{y^2-\beta y}{b^2}=0 \\\\ & \text { or } \frac{(x-\alpha / 2)^2}{a^2}-\frac{(y-\beta / 2)^2}{b^2}=\frac{1}{4}\left(\frac{\alpha^2}{a^2}-\frac{\beta^2}{b^2}\right) \end{aligned} }

Hence, centre of the hyperbola is \mathrm{\left(\frac{\alpha}{2}, \frac{\beta}{2}\right).}

Posted by

Irshad Anwar

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

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question