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)

$(-\frac{9}{2\sqrt{2}},-\frac{1}{\sqrt{2}})$

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


$\frac{2}{\frac{x_1}{9}}=\frac{-1}{\frac{-y_1}{4}}=\frac{\pm \sqrt{32}}{1}\Rightarrow x_1=\frac{9}{2\sqrt{2}}and{y}_1=\frac{1}{\sqrt{2}}Or{x}_1=-\frac{9}{2\sqrt{2}}and{y}_1=-\frac{1}{\sqrt{2}}$