Tangents are drawn to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$, parallel to the straight line $2…

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
  1. $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
  2. $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
  3. $(3 \sqrt{3},-2 \sqrt{2})$
  4. $(-3 \sqrt{3}, 2 \sqrt{2})$

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

Asked in: JEE Advanced 2012 (Paper 1)

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