Is Hyperbola part of the EAPCET Mathematics syllabus?

Yes. Hyperbola is taught in Intermediate second year · Paper IIB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Mathematics questions are there in EAPCET?

80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.

What mistakes do students make in Hyperbola?

Using the ellipse relation b² = a²(1 − e²); for a hyperbola b² = a²(e² − 1), which gives e > 1. Writing the asymptotes of x²/a² − y²/b² = 1 as y = ±(a/b)x; they are y = ±(b/a)x. Using c² = a²m² + b² for tangency; for the hyperbola the condition is c² = a²m² − b². Giving ae as the distance between the foci; that is the distance from the centre to one focus, and the distance between them is 2ae.

EAPCET Mathematics · Intermediate second year · Paper IIB

Hyperbola

Hyperbola mirrors the ellipse with key sign changes, and adds asymptotes and the rectangular hyperbola.

Subject: Mathematics
Taught in: Intermediate second year · Paper IIB
Mathematics in the paper: 80 of 160 questions

Key formulas and results

$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$ : $b^{2} = a^{2}(e^{2} - 1)$
Foci $(\pm ae, 0)$ ; asymptotes $y = \pm\dfrac{b}{a}x$
Latus rectum length $= \dfrac{2b^{2}}{a}$
$y = mx + c$ touches the hyperbola iff $c^{2} = a^{2}m^{2} - b^{2}$
Rectangular hyperbola: $a = b$ and $e = \sqrt{2}$
Difference of the focal distances of any point $= 2a$
Tangent to $\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$ at $(x_{1}, y_{1})$ : $\dfrac{xx_{1}}{a^{2}} - \dfrac{yy_{1}}{b^{2}} = 1$
Tangent at $(a\sec\theta, b\tan\theta)$ : $\dfrac{x\sec\theta}{a} - \dfrac{y\tan\theta}{b} = 1$
Normal at $(x_{1}, y_{1})$ : $\dfrac{a^{2}x}{x_{1}} + \dfrac{b^{2}y}{y_{1}} = a^{2} + b^{2}$
Normal at $(a\sec\theta, b\tan\theta)$ : $\dfrac{ax}{\sec\theta} + \dfrac{by}{\tan\theta} = a^{2} + b^{2}$

What this chapter covers

Standard equation
Eccentricity, foci and directrices
Asymptotes
Conjugate and rectangular hyperbolas
Tangents

How it is usually asked

Find the eccentricity, foci or asymptotes
Apply the tangent condition
Questions on the conjugate hyperbola

Understanding Hyperbola

A hyperbola is the set of points whose distance from a focus is e times the distance from the matching directrix, with e > 1. Equivalently, the difference of the distances from the two foci is constant, equal to 2a. In standard form x²/a² − y²/b² = 1 it opens left and right, with vertices (±a, 0), foci (±ae, 0), directrices x = ±a/e and b² = a²(e² − 1).

Most formulas match the ellipse with a sign change on b², so learning them side by side saves effort. The latus rectum is again 2b²/a, the tangent at (x₁, y₁) is xx₁/a² − yy₁/b² = 1, and y = mx + c touches the hyperbola when c² = a²m² − b². The new feature is the pair of asymptotes y = ±(b/a)x, lines the curve approaches but never meets.

The conjugate hyperbola −x²/a² + y²/b² = 1 shares the asymptotes, and the eccentricities e₁, e₂ of a hyperbola and its conjugate satisfy 1/e₁² + 1/e₂² = 1. When a = b the hyperbola is rectangular, its asymptotes are perpendicular and e = √2. EAPCET questions ask for eccentricity, foci, asymptotes, tangent conditions and conjugate-hyperbola facts.

Worked example

EAPCET-style question

For the hyperbola $9x^{2} - 16y^{2} = 144$ , the eccentricity and the distance between the foci are

(1) $e = \tfrac{5}{4}$ , distance between foci $= 10$
(2) $e = \tfrac{5}{4}$ , distance between foci $= 5$
(3) $e = \tfrac{\sqrt{7}}{4}$ , distance between foci $= 2\sqrt{7}$
(4) $e = \tfrac{5}{3}$ , distance between foci $= 10$

Solution

Divide by 144: $\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$ , so $a = 4$ and $b = 3$ .
$b^{2} = a^{2}(e^{2} - 1)$ gives $9 = 16(e^{2} - 1)$ , so $e^{2} = \tfrac{25}{16}$ and $e = \tfrac{5}{4}$ .
The foci are $(\pm ae, 0) = (\pm 5, 0)$ .
Distance between the foci $= 2ae = 10$ .

Answer: $e = \tfrac{5}{4}$ , distance between foci $= 10$

Shortcuts and results to remember

For any conic, $S_{1} = 0$ is the tangent at a point on the curve and the chord of contact from an outside point; $S_{1} = S_{11}$ is the chord with a given midpoint; $S_{1}^{2} = S\,S_{11}$ is the pair of tangents.

Common mistakes to avoid

Using the ellipse relation b² = a²(1 − e²); for a hyperbola b² = a²(e² − 1), which gives e > 1.
Writing the asymptotes of x²/a² − y²/b² = 1 as y = ±(a/b)x; they are y = ±(b/a)x.
Using c² = a²m² + b² for tangency; for the hyperbola the condition is c² = a²m² − b².
Giving ae as the distance between the foci; that is the distance from the centre to one focus, and the distance between them is 2ae.

How to prepare it

Learn the hyperbola as "ellipse with sign changes" — it halves what you need to memorise.

Once the formulas feel familiar, test them the way the exam does: practise Hyperbola questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Hyperbola chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Hyperbola part of the EAPCET Mathematics syllabus?: Yes. Hyperbola is taught in Intermediate second year · Paper IIB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Hyperbola for EAPCET?: Learn the hyperbola as "ellipse with sign changes" — it halves what you need to memorise.
How many Mathematics questions are there in EAPCET?: 80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Hyperbola?: Using the ellipse relation b² = a²(1 − e²); for a hyperbola b² = a²(e² − 1), which gives e > 1. Writing the asymptotes of x²/a² − y²/b² = 1 as y = ±(a/b)x; they are y = ±(b/a)x. Using c² = a²m² + b² for tangency; for the hyperbola the condition is c² = a²m² − b². Giving ae as the distance between the foci; that is the distance from the centre to one focus, and the distance between them is 2ae.