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 contact of the tangents on the hyperbola are

• A. $(\frac{9}{\sqrt{2}},\frac{1}{\sqrt{2}})$
• 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

The correct option is B $(-\frac{9}{2\sqrt{2}},-\frac{1}{\sqrt{2}})$

Let the point of contact of tangents on the hyperbola have coordinates $(x_1,y_1)$.

So, the equation of a tangent is

$\frac{x{x}_1}{9}-\frac{y{y}_1}{4}=1.....\left(1\right)$

The slope of the required tangent to the hyperbola is $2$.

For $y=mx+c$ to be a tangent to the hyperbola

$c^2=a^2{m}^2-b^2$

$\Rightarrow c=\pm \sqrt{9\times 4-4}$

$=\pm \sqrt{32}=\pm 4\sqrt{2}$

Hence, equations of the tangents are

$y=2x\pm 4\sqrt{2}$

$\Rightarrow 2x-y=\pm 4\sqrt{2}.....\left(2\right)$

Equating (1) and (2) we get

$\frac{x_1}{2\times 9}=\frac{y_1}{4\times 1}=\frac{1}{\pm 4\sqrt{2}}$

$\Rightarrow x_1=\pm \frac{9}{2\sqrt{2}},y_1=\pm \frac{1}{\sqrt{2}}$

B is correct.