Asymptotes and Rectangular Hyperbola

Hello JEE aspirants! Asymptotes and rectangular hyperbolas are crucial for mastering coordinate geometry. These concepts often appear in JEE Main, and understanding them well can significantly boost your score. Let's dive in!

What are Asymptotes?

Imagine a hyperbola stretching out towards infinity. Asymptotes are lines that the hyperbola approaches but never quite touches as it extends infinitely. They provide a "guide" for the hyperbola's path, showing its ultimate direction.

Geometric Intuition: Visualize the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ . As $x$ and $y$ become very large, the constant '1' becomes insignificant. The equation then approximates to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$ , which represents the combined equation of the asymptotes.

Asymptotes: $y = \pm \frac{b}{a}x \text{ or } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$

Explanation: The asymptotes are two straight lines passing through the center of the hyperbola. The slopes of these lines are $\frac{b}{a}$ and $-\frac{b}{a}$ . They intersect at the center of the hyperbola.

Rectangular Hyperbola

A special type of hyperbola where the lengths of the transverse and conjugate axes are equal (i.e., $a = b$ ) is called a rectangular hyperbola. In this case, the asymptotes are perpendicular to each other, hence the name "rectangular".

Rectangular Hyperbola: $a = b, e = \sqrt{2}$

Explanation: When $a = b$ , the eccentricity $e$ of the hyperbola is $\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2}$ . The asymptotes are $y = \pm x$ , which are clearly perpendicular.

The Equation xy = c²

Another standard form of the rectangular hyperbola is $xy = c^2$ . This equation represents a hyperbola with the coordinate axes as its asymptotes.

Rectangular Hyperbola: $xy = c^2 \text{ (eccentricity } e = \sqrt{2} \text{)}$

Derivation: This form can be obtained by rotating the standard rectangular hyperbola by 45 degrees. The key point is that the product of the coordinates of any point on the hyperbola remains constant.

Conjugate Hyperbola

The conjugate hyperbola is formed by interchanging the roles of the transverse and conjugate axes. For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , the conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ .

Conjugate Hyperbola: $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$

Geometric Meaning: The conjugate hyperbola shares the same asymptotes as the original hyperbola, but its vertices lie along the y-axis instead of the x-axis.

Relationship Between Eccentricities of Conjugate Pair

There's a beautiful relationship between the eccentricities of a hyperbola and its conjugate.

$\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \text{ (for conjugate pair)}$

Derivation: Let $e_1$ be the eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , so $e_1^2 = 1 + \frac{b^2}{a^2}$ . Let $e_2$ be the eccentricity of its conjugate $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ , so $e_2^2 = 1 + \frac{a^2}{b^2}$ . Therefore, $\frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{1}{1 + \frac{b^2}{a^2}} + \frac{1}{1 + \frac{a^2}{b^2}} = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1$ .

Hyperbola - Asymptotes = Conjugate - Asymptotes

Explanation: If you know the equation of hyperbola, you can find its asymptotes. Subtraction of hyperbola and asymptotes equation results in conjugate hyperbola equation.

Tip: Remember that the asymptotes of a hyperbola always pass through the center of the hyperbola. This is a useful fact for quickly determining the equation of the asymptotes if you know the center and the slopes.

Common Mistake: Confusing the equations of the hyperbola and its conjugate. Pay close attention to which term is positive and which is negative.

JEE Trick: When dealing with rectangular hyperbolas, remember that $xy = c^2$ is a rotated version. Look for transformations that might simplify the problem into this form.