Get Answers to all your Questions

header-bg qa

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

Option: 1

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


Option: 4

\mathrm{\text { none of these }}


Answers (1)

best_answer

Let the required point be \mathrm{P\left(x_1, y_1\right)} which is the point of intersection of tangents drawn at A and B. Then, AB is the chord of contact of point P for the hyperbola.

However, the equation of the chord of contact is \mathrm{T=0.}

or \mathrm{\quad \frac{x x_1}{a^2}-\frac{y y_1}{b^2}-1=0}                              ...........(1)

It is same as \mathrm{l x+m y+n=0}                  ...........(2)

On comparing coefficients in equation (1) and equation (2), we obtain

\mathrm{\begin{array}{ll} & \frac{x_1 / a^2}{l}=\frac{-y_1 / b^2}{m}=\frac{-1}{n} \\\\ \text { Or } \quad & x_1=-\frac{l}{a^2 n}, \quad y_1=\frac{b^2 m}{n} \\\\ \therefore \quad & \text { The required point is }\left(-\frac{l}{a^2 n}, \frac{b^2 m}{n}\right) . \end{array}}

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

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