Hyperbola Ques 36
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 mid-point of the chord of contact.
(2005, 4M)
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Answer:
Correct Answer: 36.$\frac{x^{2}}{9}-\frac{y^{2}}{4}=\frac{\left(x^{2}+y^{2}\right)^{2}}{81}$
Solution:
- Let any point on the hyperbola is $(3 \sec \theta, 2 \tan \theta)$.
$\therefore$ Chord of contact of the circle $x^{2}+y^{2}=9$ with respect to the point $(3 \sec \theta, 2 \tan \theta)$ is,
$ (3 \sec \theta) x+(2 \tan \theta) y=9 $$\quad$ …….(i)
Let $\left(x _1, y _1\right)$ be the mid-point of the chord of contact.
$\Rightarrow$ Equation of chord in mid-point form is
$ x x _1+y y _1=x _1^{2}+y _1^{2} $$\quad$ …….(ii)
Since, Eqs. (i) and (ii) are identically equal.
$\therefore \frac{3 \sec \theta}{x _1} =\frac{2 \tan \theta}{y _1} $
$ =\frac{9}{x _1^{2}+y _1^{2}} $
$\Rightarrow \sec \theta =\frac{9 x _1}{3\left(x _1^{2}+y _1^{2}\right)} $
$\text { and } \tan \theta =\frac{9 y _1}{2\left(x _1^{2}+y _1^{2}\right)}$
Thus, eliminating ’ $\theta$ ’ from above equation, we get
$ \frac{81 x _1^{2}}{9\left(x _1^{2}+y _1^{2}\right)^{2}}-\frac{81 y _1^{2}}{4\left(x _1^{2}+y _1^{2}\right)^{2}}=1 $
$ \left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right] $
$\therefore$ Required locus is $\frac{x^{2}}{9}-\frac{y^{2}}{4}=\frac{\left(x^{2}+y^{2}\right)^{2}}{81}$.
BETA
