∫ SJ Maths Mathematics Notes
UP PGT Tracker UP TGT Tracker
Coordinate Geometry

Standard equation of hyperbola

A hyperbola is the locus of a point in a plane such that the ratio of its distance from a fixed point (focus) to its fixed line (directrix) is a constant eccentricity $e > 1$.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
📖 Topic progress
○ Mark as complete

Definition and statement

Standard Equation and Fundamental Locus Principle of a Hyperbola
The standard equation of a hyperbola centered at the origin with its transverse axis along the x-axis is given by $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, where $b^2 = a^2(e^2 - 1)$ and $e > 1$ represents the eccentricity.

📐 Geometric Interpretation

The curve consists of two symmetric, unbounded branches opening left and right. It possesses two foci at $S(ae, 0)$ and $S'(-ae, 0)$, two directrices given by $x = \pm \frac{a}{e}$, and a fixed constant difference of focal distances equal to the transverse axis length $2a$, i.e., $||PS| - |PS'|| = 2a$.

⚙️ Algebraic & Structural Meaning

Represented as a second-degree homogeneous and non-homogeneous conic equation where the discriminant condition $\Delta \neq 0$ and $h^2 > ab$ holds, characterizing it as a central conic with two distinct real asymptotes given by $\frac{x}{a} \pm \frac{y}{b} = 0$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Eccentricity Bound ($e > 1$) Open interval $(1, \infty)$ Strictly Mandatory If $e = 1$, the conic degenerates into a parabola; if $0 < e < 1$, it becomes an ellipse; if $e = 0$, a circle.
Axis Orientation & Parameter Positivity ($a^2 > 0, b^2 > 0$) Real domain $\mathbb{R} \setminus (-a, a)$ for $x$ Strictly Mandatory If the signs of $a^2$ and $b^2$ are misconfigured or equal, the curve fails to represent a non-degenerate central hyperbola.
Non-Degeneracy Condition ($\Delta \neq 0$) Determinant condition of general second-degree equation Essential for conclusion If the determinant vanishes, the equation factorizes into a pair of intersecting straight lines (asymptotes).

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Eccentricity Relation for Standard Hyperbola

$$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2}{a^2}}$$

Rule: For a standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, use the direct coefficient ratio to compute eccentricity instantly without solving auxiliary equations.

Application: Frequently asked in UP PGT to find $e$ given $a$ and $b$ or vice-versa.
02

Equation of Asymptotes

$$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 $$

Rule: To find the combined equation of asymptotes, replace the constant $1$ on the RHS with $0$. For separate asymptotes, factorize the resulting expression.

03

Conjugate Hyperbola Eccentricity Relation

$$e_1^{-2} + e_2^{-2} = 1$$

Rule: If $e_1$ and $e_2$ are eccentricities of a hyperbola and its conjugate hyperbola, their respective eccentricities satisfy the invariant relation $e_1^{-2} + e_2^{-2} = 1$.

Application: Directly tested in UP TGT/PGT to calculate one eccentricity when the other is known.
04

Latus Rectum Length & Coordinates

$$ \text{Latus Rectum} = \frac{2b^2}{a}, \quad \text{Endpoints: } \left(\pm ae, \pm \frac{b^2}{a}\right) $$

Rule: Memorize the latus rectum formula directly to bypass deriving focal chords.

Application: Crucial for numerical problems involving length and focal properties.
05

Director Circle Radius

$$ x^2 + y^2 = a^2 - b^2 $$

Rule: The locus of the point of intersection of perpendicular tangents is a circle. If $a^2 \le b^2$, the director circle is imaginary or a point circle.

Application: Tested in elite recruitment papers (UP PGT & LT Grade) to check condition for real director circles ($a > b$).
06

Rapid Option Elimination via Asymptote Slopes

$$ m = \pm \frac{b}{a} $$

Rule: Check the slopes of the given options against $\pm \frac{b}{a}$ to eliminate two incorrect options instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Transverse and Conjugate Axes

Common Pitfall: Assuming $a$ is always the denominator of the $x^2$ term. In vertical hyperbolas, the $y^2$ term has the positive coefficient.

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

Why it fails: Using $a$ for the x-axis when the transverse axis lies along the y-axis flips the foci coordinates to $(0, \pm ae)$ and directrices to $y = \pm \frac{a}{e}$.

🛡️ Exam Defense: Always identify the positive term in the standard form $\frac{\text{Positive}}{\dots} - \frac{\text{Negative}}{\dots} = 1$ to designate the transverse axis.
⚠️ COMMON ERROR #02

Rectangular Hyperbola Eccentricity Confusion

Common Pitfall: Applying general eccentricity formulas to a rectangular hyperbola without recalling its unique property.

Counterexample: $a = b \implies e = \sqrt{2}$

Why it fails: For a rectangular (or equilateral) hyperbola, the semi-transverse and semi-conjugate axes are equal ($a = b$), fixing the eccentricity permanently at $\sqrt{2}$.

🛡️ Exam Defense: Recognize $xy = c^2$ or $x^2 - y^2 = a^2$ as rectangular hyperbolas and immediately substitute $e = \sqrt{2}$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Standard Horizontal Hyperbola$a > 0, b > 0, e > 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1Transverse axis along x-axisPositive term on $x^2$
Standard Vertical Hyperbola$a > 0, b > 0, e > 1$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1Transverse axis along y-axisPositive term on $y^2$
Rectangular Hyperbola$a = b$x^2 - y^2 = a^2 \text{ or } xy = c^2Asymptotes intersect at 90^\circMention of 'rectangular' or $e = \sqrt{2}$
Conjugate HyperbolaInterchanged axes\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1Axes swapped with original hyperbolaNegative sign on standard form

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the eccentricity of the hyperbola $9x^2 - 16y^2 = 144$.

(A) \frac{5}{4}(B) \frac{4}{3}(C) \frac{5}{3}(D) \frac{3}{2}
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Rewrite equation in standard form by dividing by $144$: $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Step 2: Identify $a^2 = 16$ and $b^2 = 9$. Step 3: Apply $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
⚡ Method B: 25-Second Exam Shortcut
Use $e = \sqrt{1 + \frac{\text{coeff of } x^2 \text{ denominator}}{\text{coeff of } y^2 \text{ denominator}}} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$ instantly.
💡 Tip: Always reduce the constant term to $1$ first before reading off $a^2$ and $b^2$.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If the distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$, find its equation.

(A) x^2 - y^2 = 32(B) x^2 - y^2 = 16(C) x^2 - y^2 = 64(D) \frac{x^2}{16} - \frac{y^2}{16} = 1
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Distance between foci $= 2ae = 16 \implies ae = 8$. Step 2: Given $e = \sqrt{2}$, so $a\sqrt{2} = 8 \implies a = \frac{8}{\sqrt{2}} = 4\sqrt{2}$. Step 3: $a^2 = 32$. Step 4: For $e = \sqrt{2}$ (rectangular hyperbola), $a^2 = b^2 = 32$. Step 5: Equation is $\frac{x^2}{32} - \frac{y^2}{32} = 1 \implies x^2 - y^2 = 32$? Wait, let's recalculate: $2ae = 16 \implies ae = 8$. $a(\sqrt{2}) = 8 \implies a = 4\sqrt{2} \implies a^2 = 32$. $b^2 = a^2(e^2 - 1) = 32(2 - 1) = 32$. Thus $\frac{x^2}{32} - \frac{y^2}{32} = 1 \implies x^2 - y^2 = 32$. Correct option is (A). Let's verify option alignment.
⚡ Method B: 25-Second Exam Shortcut
Notice $e = \sqrt{2} \implies$ rectangular hyperbola $x^2 - y^2 = a^2$. Focal distance $2ae = 16 \implies 2a\sqrt{2} = 16 \implies a = 4\sqrt{2} \implies a^2 = 32$. Hence $x^2 - y^2 = 32$.
💡 Tip: For rectangular hyperbola, recognize immediately that $2ae = 16 \implies a = \frac{8}{\sqrt{2}}$.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The eccentricity of the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ is:
A $\frac{4}{3}$
B $\frac{5}{3}$
C $\frac{5}{4}$
D $\frac{7}{4}$
Derivation & Analysis:

$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.

Q2
MCQ 2: The foci of the hyperbola $16x^2 - 9y^2 = 144$ are:
A $(\pm 5, 0)$
B $(0, \pm 5)$
C $(\pm 4, 0)$
D $(\pm 3, 0)$
Derivation & Analysis:

Equation reduces to $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Here $a^2 = 9, b^2 = 16$, $e = \frac{5}{3}$. Foci are $(\pm ae, 0) = \left(\pm 3 \cdot \frac{5}{3}, 0\right) = (\pm 5, 0)$.

Q3
MCQ 3: If the eccentricity of a hyperbola is $2$, find the angle between its asymptotes.
A $\frac{\pi}{3}$
B $\frac{\pi}{2}$
C $2\sec^{-1}(2)$
D $2\tan^{-1}(\sqrt{3})$
Derivation & Analysis:

Slopes of asymptotes are $m_1 = \frac{b}{a}, m_2 = -\frac{b}{a}$. Since $e = 2 \implies \frac{b}{a} = \sqrt{e^2 - 1} = \sqrt{3}$. Angle $\theta = 2\tan^{-1}\left(\frac{b}{a}\right) = 2\tan^{-1}(\sqrt{3}) = \frac{2\pi}{3}$.

Q4
MCQ 4: The equation of the directrices of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are:
A $x = \pm \frac{a}{e}$
B $y = \pm \frac{a}{e}$
C $x = \pm ae$
D $y = \pm \frac{e}{a}$
Derivation & Analysis:

By standard definition, the directrices for a horizontal hyperbola are vertical lines given by $x = \pm \frac{a}{e}$.

Q5
MCQ 5: The product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ to its asymptotes is:
A $\frac{a^2 b^2}{a^2 + b^2}$
B $\frac{a^2 + b^2}{a^2 b^2}$
C ab
D $a^2$ + $b^2$
Derivation & Analysis:

The fixed product of perpendicular distances from any point on a hyperbola to its asymptotes is constantly equal to $\frac{a^2 b^2}{a^2 + b^2}$.

Q6
MCQ 6: If $e_1$ and $e_2$ are the eccentricities of a hyperbola and its conjugate hyperbola, then:
A $e_1^2 + e_2^2 = 1$
B $e_1^{-2} + e_2^{-2} = 1$
C $e_1 = e_2$
D $e_1 e_2 = 1$
Derivation & Analysis:

For conjugate hyperbolas, $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$ is a standard identity derived from $e_1^2 = 1 + \frac{b^2}{a^2}$ and $e_2^2 = 1 + \frac{a^2}{b^2}$.

Q7
MCQ 7: The length of the latus rectum of the hyperbola $9x^2 - 16y^2 = 144$ is:
A $\frac{9}{2}$
B $\frac{16}{3}$
C 9
D $\frac{8}{3}$
Derivation & Analysis:

Here $a^2 = 16 \implies a = 4$, and $b^2 = 9 \implies b = 3$. Latus rectum = $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = \frac{9}{2}$.

Q8
MCQ 8: The locus of the point of intersection of perpendicular tangents to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (where $a > b$) is:
A $x^2$ + $y^2$ = $a^2$ + $b^2$
B $x^2$ + $y^2$ = $a^2$ - $b^2$
C $x^2$ + $y^2$ = $b^2$ - $a^2$
D $x^2$ + $y^2$ = $a^2$
Derivation & Analysis:

The director circle of a hyperbola is given by $x^2 + y^2 = a^2 - b^2$, which is real only when $a > b$.

Q9
MCQ 9: For what value of $k$ does the equation $x^2 - y^2 - 4x + 4y + k = 0$ represent a pair of straight lines?
A $k = 0$
B $k = 4$
C $k = -4$
D Any real number
Derivation & Analysis:

Completing squares: $(x-2)^2 - (y-2)^2 = 4 - k$. For a degenerate conic (pair of intersecting lines), the RHS must be $0$, giving $k = 4$? Wait, $(x-2)^2 - (y-2)^2 = 4-k$. If $k=4$, RHS is 0, so $(x-2)^2 = (y-2)^2 \implies y-2 = \pm(x-2)$, which is a pair of lines. Thus $k = 4$.

Q10
MCQ 10: The eccentricity of the rectangular hyperbola is always:
A 1
B $\sqrt{2}$
C 2
D $\sqrt{3}$
Derivation & Analysis:

For a rectangular hyperbola, $a = b$, so $e = \sqrt{1 + \frac{a^2}{a^2}} = \sqrt{2}$.

Common questions

Both have equations of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. If $h^2 < ab$, it represents an ellipse (or circle); if $h^2 > ab$, it represents a hyperbola; and if $h^2 = ab$, a parabola.

No. By definition, eccentricity $e$ for a hyperbola must strictly satisfy $e > 1$. If $e = 1$, it is a parabola; if $e < 1$, it is an ellipse.

Memorize the direct shortcuts: substituting $0$ for the constant term to get asymptotes, and using $e = \sqrt{1 + \frac{b^2}{a^2}}$ for standard forms to bypass lengthy derivations.

← Previous Topic
Standard equation of ellipse
Next Topic →
Condition for a Line to be Normal to a Conic