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

Answer 1

The Correct Option is B

Equation of tangent to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$y=mx\pm\sqrt{a^2m^2-b^2}$
Description of Situation$ \, \, \, \, $ If two straight lines
$\, \, \, \, \, \, \, \, \, \, \, \, a_1x+b_1y+c_1=0$
and $a_2x+b_2y+c_2=0\, \, \, are identical.Then,\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c-1}{c-2}$
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 (x_1,y_1) is
$\, \, \, \, \, \, \, \, \frac{xx_1}{9} - \frac{yy_1}{4}=1\, \, \, \, \, \, \, \, \, \, \, \, \, .....(ii)$
From Eqs. (i) and (ii),
$\frac{2}{\frac{x_1}{9}}=\frac{-1}{\frac{-y_1}{4}}=\frac{\pm\sqrt{32}}{1} \Rightarrow x_1=-\frac{9}{2\sqrt2} and y_1=-\frac{1}{\sqrt2}$
or$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, x_1=\frac{9}
{2\sqrt2},y_1=\frac{1}{\sqrt2}$