Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The point of intersection of tangents drawn to 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%

The point of intersection of tangents drawn to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}  at the points where it is intersected by the line \mathrm{I x+m y+n=0} is

Option: 1

\mathrm{\left(\frac{-a^2 I}{n}, \frac{b^2 m}{n}\right)}


Option: 2

\mathrm{\left(\frac{a^2 l}{n}, \frac{-b^2 m}{n}\right)}


Option: 3

\mathrm{\left(-\frac{a^2 n}{l}, \frac{b^2 n}{m}\right)}


Option: 4

\mathrm{\left(\frac{a^2 n}{l}, \frac{-b^2 n}{m}\right)}


Answers (1)

best_answer

Let \mathrm{(x_{1},y_{1})} be the required point. Then the equation of the chord of contact of tangents drawn from \mathrm{(x_{1},y_{1})} to the given hyperbola is
 \mathrm{\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1\ \ \ ............(i)}
The given line is \mathrm{l x+m y+n=0\ \ \ ............(ii)}
Equation \mathrm{(i)} and \mathrm{(ii)} represent the same line
\mathrm{\therefore \quad \frac{x_1}{a^2 l}=-\frac{y_1}{b^2 m}=\frac{1}{-h} \Rightarrow x_1=\frac{-a^2 l}{n}, y_1=\frac{b^2 m}{n}} ; Hence the required point is \mathrm{\left(-\frac{a^2 l}{n}, \frac{b^2 m}{n}\right)}

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question