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

Answer 1

Answer: (A), (B)

PLAN Equation of tangent to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$y=mx\pm \sqrt{a^2{m}^2-b^2}$
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 $y=2x-1$ is
$y=2x\pm \sqrt{9(4)-4}$
$\therefore y=2x\pm \sqrt{32}....$(i)
The equation of tangent at $\left(x_1,y_1\right)$ is
$\frac{x{x}_1}{9}-\frac{y{y}_1}{4}=\mathrm{1....}$(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 $x_1=\frac{9}{2\sqrt{2}},y_1=\frac{1}{\sqrt{2}}$