Question

Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, parallel to the straight line $2x-y=1.$ The points of contacts of the tangents on the hyperbola are

• A. $(\frac{9}{2\sqrt2},\frac{1}{\sqrt2})$
• B. $(-\frac{9}{2\sqrt2},\frac{1}{\sqrt2})$
• C. $(3\sqrt3,-2\sqrt2)$
• D. $(-3\sqrt3,2\sqrt2)$

Answer 1

Answer: (A), (B)

PLAN Equation of tangent to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is
$y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$
Description of Situation If two straight lines
$a_{1} x+b_{1} y+c_{1}=0$
and $a_{2} x+b_{2} y+c_{2}=0$ are identical.
Then, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Equation of tangent, parallel to $y=2 x-1$ is
$ y=2 x \pm \sqrt{9(4)-4} $
$\therefore y=2 x \pm \sqrt{32} ....$(i)
The equation of tangent at $\left(x_{1}, y_{1}\right)$ is
$\frac{x x_{1}}{9}-\frac{y y_{1}}{4}=1 ....$(ii)
From Eqs. (i) and (ii),

$\frac{2}{\frac{x_{1}}{9}}=\frac{-1}{\frac{-y_{1}}{4}}=\frac{\pm \sqrt{32}}{1} $
$\Rightarrow x_{1}=-\frac{9}{2 \sqrt{2}}$ and $y_{1}=-\frac{1}{\sqrt{2}}$
or $ x_{1}=\frac{9}{2 \sqrt{2}}, y_{1}=\frac{1}{\sqrt{2}}$