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 mid-points of the chords of the circle  wh

The locus of the mid-points of the chords of the circle \mathrm{x^2+y^2=16} which are tangent to the hyperbola \mathrm{9 x^2-16 y^2=144} is

Option: 1

\mathrm{\left(x^2+y^2\right)^2=16 x^2-9 y^2}


Option: 2

\mathrm{\left(x^2+y^2\right)^2=9 x^2-16 y^2}


Option: 3

\mathrm{\left(x^2-y^2\right)^2=16 x^2-9 y^2}


Option: 4

None of these


Answers (1)

best_answer

The given hyperbola is \mathrm{\frac{x^2}{16}-\frac{y^2}{9}=1\ \ \ ............(i)}
Any tangent to \mathrm{(i)} is \mathrm{y=m x+\sqrt{16 m^2-9}\ \ \ ............(ii)}
Let \mathrm{(x_{1},y_{1})} be the mid point of the chord of the circle \mathrm{x^2+y^2=16}
Then equation of the chord is \mathrm{T=S_1 \text { i.e., } x x_1+y y_1-\left(x_1^2+y_1^2\right)=0\ \ \ ..........(iii)}
Since \mathrm{(ii)} and \mathrm{(iii)} represent the same line.
\mathrm{\therefore \quad \frac{m}{x_1}=\frac{-1}{y_1}=\frac{\sqrt{16 m^2-9}}{-\left(x_1^2+y_1^2\right)}}
\mathrm{\Rightarrow \quad m=-\frac{x_1}{y_1} \quad \text { and } \quad\left(x_1^2+y_1^2\right)^2=y_1^2\left(16 m^2-9\right) \quad \Rightarrow \quad\left(x_1^2+y_1^2\right)^2=16 \cdot \frac{x_1^2}{y_1^2} y_1^2-9 y_1^2=}
\mathrm{16 x_1^2-9 y_1^2}
\mathrm{\therefore } Locus of \mathrm{\left(x_1, y_1\right) \text { is }\left(x^2+y^2\right)^2=16 x^2-9 y^2}

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