Question

Tangents are drawn from any point on the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}$ =1 to the circle $x^2+y^2=9$ Find the locus of midpoint of the chord of contact

Answer 1

Any point on the hyperbola
$\frac{x^2}{9}-\frac{y^2}{4}=1$ to the circle $x^2+y^2=9$ =1 is $(3sec\theta ,2tan\theta ).$
Chord of contact of the circle $x^2+y^2=9$with respect to the point $(3sec\theta ,2tan\theta ).$ is $3sec\theta x+2tan\theta y=9$ ...(1)
Let $(x_1,y_1)$ be the mid-point of the chord of contact.
$=⟩$ equation of chord in mid-point form is $x{x}_1+y{y}_1=x_1^2+y_1^2$ (2)
Since (1) and (2) represent the same line,
$\frac{x_1}{3\sec \theta }=\frac{y_1}{2\tan \theta }=\frac{{x_1}^2+{y_1}^2}{9}$
$=⟩\sec \theta =\frac{9x_1}{3(x_1^2+y_{1^2})}$ , $\tan \theta =\frac{9y_1}{2(x_1^2+y_{1^2})}$
Hence $\frac{81x_1^2}{9(x_1^2+y_{1^2}{)}^2}-\frac{81y_{1^2}}{4(x_1^2+y_{1^2}{)}^2}=1$
$=⟩$ the required locus is $\frac{x^2}{9}-\frac{y^2}{4}=(\frac{x^2+y^2}{9}{)}^2$