Tangents are drawn to the hyperbola x²/9 - y²/4 = 1, parallel to the straight line 2x - y = 1. The p

Q: Why are there two points of contact instead of just one?

The hyperbola consists of two separate branches. A line with a given slope can be tangent to the right branch and the left branch simultaneously. Because the hyperbola is symmetric about the origin, the points of contact are also symmetric.

Q: Can I use calculus to find the point of contact?

Absolutely! You could differentiate the hyperbola equation to find the slope of the tangent at any point $(x_1, y_1)$, set it equal to 2, and solve the system of equations. However, the formula approach is much faster for JEE exams.

Q: What if the value inside the square root for $c$ is negative?

That would mean the line cannot be tangent to the hyperbola. In such a case, the line would either intersect the hyperbola at two points (secant) or not intersect it at all. For a tangent to exist, we must have $a^2m^2 > b^2$.

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

Options:

(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) $(-3\sqrt{3}, 2\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The hyperbola is defined by the equation:

$$\frac{x^2}{9} - \frac{y^2}{4} = 1$$

By comparing this to the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we identify the parameters $a^2 = 9$ and $b^2 = 4$.

The given line is $2x - y = 1$. Rewriting this in slope-intercept form $y = mx + c$, we obtain $y = 2x - 1$. Thus, the slope of the tangent is $m = 2$.

### The Master Equation

For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the point of contact $(x_1, y_1)$ for a tangent with slope $m$ is given by the coordinates:

$$\left( \pm \frac{a^2m}{\sqrt{a^2m^2 - b^2}}, \pm \frac{b^2}{\sqrt{a^2m^2 - b^2}} ight)$$

This formula allows us to bypass the construction of a quadratic equation by discriminant analysis, saving valuable time during examinations.

### The Calculation

First, we calculate the radical term $c = \sqrt{a^2m^2 - b^2}$. Substituting our known values:

$$c = \sqrt{9(2)^2 - 4} = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$$

Now, we determine the $x$-coordinate:

$$x = \pm \frac{a^2m}{c} = \pm \frac{9 	imes 2}{4\sqrt{2}} = \pm \frac{18}{4\sqrt{2}} = \pm \frac{9}{2\sqrt{2}}$$

Next, we determine the $y$-coordinate:

$$y = \pm \frac{b^2}{c} = \pm \frac{4}{4\sqrt{2}} = \pm \frac{1}{\sqrt{2}}$$

### Final Result

The points of contact where the line with slope $2$ is tangent to the hyperbola are:

$$\mathbf{\left( \frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}} ight) 	ext{ and } \left( -\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}} ight)}$$

These points exhibit perfect symmetry about the origin, reflecting the inherent geometric properties of the hyperbola. You have successfully navigated the derivation through logical, systematic steps.

Animated Solution for Mathematics - Conic Sections: 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

Select Answer:

Visualized Solution

Visualizing the Hyperbola and Line

Identifying Parameters $a^{2}$ , $b^{2}$ , and $m$

The Point of Contact Formula

Calculating the Denominator $a^{2} m^{2} - b^{2}$

Evaluating the Denominator

Substituting into the Coordinates

Final Simplification

Conclusion and Final Points

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Master Equation

The Calculation

Final Result

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