Q31 / 60 · JEE Advanced 2012 (Paper 1) · Mathematics
Tangents are drawn to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$, parallel to the straight line $2 x-y=1$. The points of contact of the tangents on the hyperbola are
- Correct answerA$\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
- Correct answerB$\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
- C$(3 \sqrt{3},-2 \sqrt{2})$
- D$(-3 \sqrt{3}, 2 \sqrt{2})$
Solution
If slope of tangents to hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $m$,
then equations of tangent to the hyperbola is
$y=m x \pm \sqrt{a^{2} m^{2}-b^{2}} \quad$ with the points of contact
$\quad\left(\quad \pm a^{2} m\right.$
$\left.\frac{\pm \sqrt{a^{2} m^{2}-b^{2}}}{\sqrt{a^{2} m^{2}-b^{2}}}\right)$
$\therefore$ Tangent to hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ is parallel to $2 x-y=1$,
$\therefore$ Slope of tangent $=2$
$\therefore$ Points of contact are $\left(\frac{\pm 9 \times 2}{\sqrt{9 \times 4-4}}, \frac{\pm 4}{\sqrt{9 \times 4-4}}\right)$
i.e. $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{-9}{2 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)$
Hence, the correct answer is options: 'A', 'B'.