Hyperbola: Eccentricity, Asymptotes and Tangents | NIMCET

Quick answer: The hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}}$ (so $e > 1$), foci $(\pm ae, 0)$, directrices $x = \pm\frac{a}{e}$, latus rectum $\frac{2b^2}{a}$ and asymptotes $y = \pm\frac{b}{a}x$. For any point on it, the difference of focal distances is $2a$. The line $y = mx + c$ is a tangent when $c^2 = a^2m^2 - b^2$.

Compare it with the ellipse: the formulas look alike, but the signs of $b^2$ change.

Standard hyperbola

Focal distances: $\lvert SP - S'P\rvert = 2a$.

Ellipse versus hyperbola

Conjugate and rectangular hyperbolas

Conjugate hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ has foci on the y-axis. If $e_1$ and $e_2$ are the eccentricities of a hyperbola and its conjugate,

$\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$

Rectangular hyperbola: $a = b$, so $x^2 - y^2 = a^2$, with $e = \sqrt{2}$ and perpendicular asymptotes $y = \pm x$. Rotated, it takes the form $xy = c^2$, with parametric point $\left(ct, \frac{c}{t}\right)$.

Tangents and normals

Worked examples

Example 1: the basics

For $\frac{x^2}{16} - \frac{y^2}{9} = 1$, find e, the foci and the asymptotes.

Step 1: $e = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$.

Step 2: Foci $(\pm 5, 0)$, asymptotes $y = \pm\frac{3}{4}x$.

Example 2: equation from data

Find the hyperbola with foci $(\pm 6, 0)$ and eccentricity 2.

Step 1: $ae = 6$, so $a = 3$.

Step 2: $b^2 = 9(4 - 1) = 27$.

Answer: $\frac{x^2}{9} - \frac{y^2}{27} = 1$.

Example 3: conjugate eccentricity

If the eccentricity of a hyperbola is $\frac{5}{4}$, find that of its conjugate.

$\frac{1}{e_2^2} = 1 - \frac{16}{25} = \frac{9}{25}$, so $e_2 = \frac{5}{3}$.

Check: using Example 1, the conjugate of $\frac{x^2}{16} - \frac{y^2}{9} = 1$ has $e_2 = \sqrt{1 + \frac{16}{9}} = \frac{5}{3}$ ✓.

Example 4: tangency

Find c so that $y = 2x + c$ touches $\frac{x^2}{4} - \frac{y^2}{3} = 1$.

$c^2 = 4 \cdot 4 - 3 = 13$, so $c = \pm\sqrt{13}$.

Summary table

Exam tips

• The positive term decides the transverse axis: $x^2$ positive means foci on the x-axis.
• Asymptotes come from replacing the 1 on the right with 0.
• Any hyperbola with perpendicular asymptotes is rectangular, so $e = \sqrt{2}$.

See all topics in the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

• Using the ellipse formula $e = \sqrt{1 - \frac{b^2}{a^2}}$.
• Writing $SP + S'P = 2a$ for a hyperbola.
• Taking asymptotes as $y = \pm\frac{a}{b}x$.
• Assuming a hyperbola must have $a > b$; it need not.

Frequently Asked Questions

What is the eccentricity of a hyperbola?

For x²/a² − y²/b² = 1, e = √(1 + b²/a²), and it is always greater than 1.

What are the asymptotes of a hyperbola?

For x²/a² − y²/b² = 1, the asymptotes are y = ±(b/a)x.

What is a rectangular hyperbola?

A hyperbola with a = b, such as x² − y² = a². Its asymptotes are perpendicular and its eccentricity is √2.

What is the condition for y = mx + c to touch a hyperbola?

c² = a²m² − b².

How are the eccentricities of a hyperbola and its conjugate related?

1/e₁² + 1/e₂² = 1.