Points of Contact of Tangents to Hyperbola Parallel to a Line

To find the points of contact of the tangents to the hyperbola that are parallel to the given line, we proceed as follows:

The given equation of the hyperbola is: $\frac{x^2}{9} - \frac{y^2}{4} = 1$ Here, $a^2 = 9$ and $b^2 = 4$.

The given straight line is: $2x - y = 1 \implies y = 2x - 1$ The slope of this line is $m = 2$. Since the tangents are parallel to this line, the slope of each tangent must also be $m = 2$.

Method 1: Using Calculus

Differentiating the equation of the hyperbola implicitly with respect to $x$: $\frac{d}{dx}\left(\frac{x^2}{9} - \frac{y^2}{4}\right) = 0$ $\frac{2x}{9} - \frac{2y}{4}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{4x}{9y}$

At the point of contact $(x_1, y_1)$, the slope of the tangent is equal to $m = 2$: $\frac{4x_1}{9y_1} = 2 \implies x_1 = \frac{9}{2}y_1$

Since the point of contact $(x_1, y_1)$ lies on the hyperbola: $\frac{x_1^2}{9} - \frac{y_1^2}{4} = 1$

Substituting $x_1 = \frac{9}{2}y_1$ into the hyperbola's equation: $\frac{\left(\frac{9}{2}y_1\right)^2}{9} - \frac{y_1^2}{4} = 1$ $\frac{81 y_1^2}{4 \times 9} - \frac{y_1^2}{4} = 1$ $\frac{9y_1^2}{4} - \frac{y_1^2}{4} = 1$ $\frac{8y_1^2}{4} = 1 \implies 2y_1^2 = 1 \implies y_1^2 = \frac{1}{2}$

Thus, the values for $y_1$ are: $y_1 = \pm \frac{1}{\sqrt{2}}$

Correspondingly, the values for $x_1$ are:

• For $y_1 = \frac{1}{\sqrt{2}}$: $x_1 = \frac{9}{2}\left(\frac{1}{\sqrt{2}}\right) = \frac{9}{2\sqrt{2}}$ This gives the point of contact: $\left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

• For $y_1 = -\frac{1}{\sqrt{2}}$: $x_1 = \frac{9}{2}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{9}{2\sqrt{2}}$ This gives the point of contact: $\left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$

Therefore, the points of contact of the tangents are: $\left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \quad \text{and} \quad \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$

These correspond to options (A) and (B).