Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A variable chord 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%7Ba%5E2%7D-%5Cfrac%7By%5E

A variable chord of the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} is a tangent to the circle \mathrm{x^2+y^2=c^2}. The locus of its middle point is

Option: 1

\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)^2=c^2\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}\right)


Option: 2

\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2=c^2\left(\frac{x^2}{a^4}-\frac{y^2}{b^4}\right)


Option: 3

\left(\frac{x^2}{a^4}-\frac{y^2}{b^4}\right)^2=c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)


Option: 4

\text { None of these }


Answers (1)

best_answer

Let \left(x_1, y_1\right) be the middle point of a variable chord of the hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.

Equation of the chord is \frac{x x_1}{a^2}-\frac{y y_1}{b^2}=\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}                       ...(i)

[Using T=S_1 ]

If this chord touches the circle x^2+y^2=c^2, then length of \perp from the centre (0,0) upon (i) is equal to the radius c.

\therefore \pm \frac{\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}}{\sqrt{\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}}}=c \Rightarrow c^2\left(\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}\right)=\left(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}\right)^2

Hence, locus of \left(x_1, y_1\right)\, \, is \, \, \left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)^2=c^2\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}\right)

Posted by

Anam Khan

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

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30
anam-khan

DASA, CSAB Special round 2 seat allotment result 2025 declared on csab.nic.in

Latest Question