Find asymptotes of x²/a² - y²/b² = 1 and prove e > 1.

Hyperbola — Asymptotes and Eccentricity

Question

For the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , find the equations of the asymptotes. Also prove that the eccentricity $e > 1$ for every hyperbola.

This is a two-part conceptual question that JEE Advanced loves. Knowing the derivation — not just memorising the result — is what separates a 90-percentiler from a 99-percentiler.

Solution — Step by Step

An asymptote is a line that the curve approaches but never touches as $x \to \pm\infty$ . For the hyperbola, we look for straight lines $y = mx + c$ such that the vertical distance between the line and the curve tends to zero.

From the equation:

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \implies y^2 = \frac{b^2}{a^2}(x^2 - a^2)

y = \pm \frac{b}{a} \sqrt{x^2 - a^2}

Factor out $x$ from the square root:

y = \pm \frac{bx}{a} \sqrt{1 - \frac{a^2}{x^2}}

As $x \to \infty$ , the term $\frac{a^2}{x^2} \to 0$ , so:

y \to \pm \frac{bx}{a} \cdot 1 = \pm \frac{b}{a}x

The curve approaches the lines $y = \frac{b}{a}x$ and $y = -\frac{b}{a}x$ . These are the asymptotes.

\boxed{y = \pm \frac{b}{a}x} \quad \text{or equivalently,} \quad \frac{x}{a} \pm \frac{y}{b} = 0

The combined equation of both asymptotes is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$ . Notice it is the hyperbola equation with the RHS set to zero — this is a trick worth remembering for quick asymptote writing.

For the standard hyperbola, the relationship between $a$, $b$, and $e$ is:

b^2 = a^2(e^2 - 1)

Since $b^2 > 0$ (b is a real positive length), we need:

a^2(e^2 - 1) > 0

Since $a^2 > 0$ always, we require $e^2 - 1 > 0$ , which gives $e^2 > 1$ , so $e > 1$ (taking the positive root). Hence proved.

Why This Works

The asymptote derivation works because we are finding the “limiting slope” of the hyperbola at infinity. The hyperbola’s branches spread outward, getting closer and closer to those two straight lines without ever meeting them.

The key insight in the eccentricity proof is that $b^2 = a^2(e^2 - 1)$ is not some arbitrary formula — it comes directly from the focus-directrix definition. Since $b$ must be a real number, $e^2 - 1$ cannot be zero or negative, forcing $e > 1$ .

This is why a hyperbola “opens up” while an ellipse closes — for an ellipse, $b^2 = a^2(1 - e^2)$ requires $e < 1$ instead.

Alternative Method

Using the condition for a line to be an asymptote directly:

Substitute $y = mx + c$ into the hyperbola equation and collect terms:

\frac{x^2}{a^2} - \frac{(mx+c)^2}{b^2} = 1

x^2\left(\frac{1}{a^2} - \frac{m^2}{b^2}\right) - \frac{2mcx}{b^2} - \frac{c^2}{b^2} - 1 = 0

For the line to be an asymptote, the curve must approach the line at infinity — meaning both the coefficient of $x^2$ AND the coefficient of $x$ must vanish (otherwise we get finite intersection points, not asymptotic behaviour).

Setting the coefficient of $x^2$ to zero: $\frac{1}{a^2} = \frac{m^2}{b^2}$ , giving $m = \pm\frac{b}{a}$ .

Setting the coefficient of $x$ to zero: $-\frac{2mc}{b^2} = 0$ , giving $c = 0$ .

So the asymptotes are $y = \pm\frac{b}{a}x$ — same answer, more rigorous derivation.

In JEE Advanced, this “two conditions” approach is often tested directly: “A line is an asymptote if it meets the conic at two points both tending to infinity.” Setting coefficient of $x^2$ = 0 handles one point; setting coefficient of $x$ = 0 handles the other.

Common Mistake

Students often write the asymptotes as $y = \pm\frac{a}{b}x$ — flipping $a$ and $b$ . Remember: the asymptote slope is $\frac{b}{a}$ , where $b$ is under the $y^2$ term. A quick sanity check: if $b > a$ , the asymptotes are steeper than 45°, which makes sense geometrically since the hyperbola opens more “narrowly.” If you get slope $< 1$ when $b > a$ , you’ve flipped it.

Also, many students forget to prove $b^2 > 0$ explicitly when showing $e > 1$ . In JEE Advanced, that one line of justification is what earns the mark.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next