IIT JEE Main Maths -Unit 10- Standard equation and parameters (center, foci, asymptotes, eccentricity)- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 10- Standard equation and parameters (center, foci, asymptotes, eccentricity) – Study Notes – New syllabus
IIT JEE Main Maths -Unit 10- Standard equation and parameters (center, foci, asymptotes, eccentricity) – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
Hyperbola: Standard Equation and Parameters
Hyperbola: Parametric Form
Hyperbola: Standard Equation and Parameters
A hyperbola is the locus of all points for which the difference of distances from two fixed points (foci) is constant.
Standard hyperbolas have two branches and possess asymptotes.
1. Standard Equation of Hyperbola
(A) Hyperbola Opening Left–Right (Horizontal Transverse Axis)
\( \dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1,\quad a>0,\ b>0 \)
(B) Hyperbola Opening Up–Down (Vertical Transverse Axis)
\( \dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1,\quad a>0,\ b>0 \)
In both cases:
- \( a = \) semi-transverse axis
- \( b = \) semi-conjugate axis
- \( c = \) distance of focus from center
Key identity for hyperbola:
\( c^2 = a^2 + b^2 \)
(This is opposite to ellipse, where 𝑐 2 = 𝑎 2 − 𝑏 2 c 2 =a 2 −b 2 .)
2. Parameters of Hyperbola
| Parameter | Horizontal Hyperbola | Vertical Hyperbola |
| Center | \( (0,0) \) | \( (0,0) \) |
| Vertices | \( (\pm a,0) \) | \( (0,\pm a) \) |
| Foci | \( (\pm c,0) \) | \( (0,\pm c) \) |
| Asymptotes | \( y = \pm\dfrac{b}{a}x \) | \( y = \pm\dfrac{a}{b}x \) |
| Eccentricity | \( e = \dfrac{c}{a} > 1 \) | \( e = \dfrac{c}{a} > 1 \) |
3. Eccentricity of Hyperbola
Eccentricity for hyperbola is always greater than 1:
\( e = \dfrac{c}{a},\quad c^2 = a^2 + b^2,\quad e>1 \)
Using identity,
\( b^2 = a^2(e^2 – 1) \)
4. Asymptotes of Hyperbola
For hyperbola:
\( \dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1 \)
Asymptotes are:
\( y = \pm\dfrac{b}{a}x \)
Similarly, for vertical hyperbola:
\( y = \pm\dfrac{a}{b}x \)
Asymptotes pass through center and bound the branches of hyperbola.
5. Conjugate Hyperbola
If the main hyperbola is:
\( \dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1, \)
its conjugate hyperbola is:
\( \dfrac{y^2}{b^2} – \dfrac{x^2}{a^2} = 1 \)
6. Parametric Coordinates
Parametric point on hyperbola:
\( (x,y) = (a\sec\theta,\ b\tan\theta) \)
Useful for tangent and normal equations.
Example
For hyperbola \( \dfrac{x^2}{9}-\dfrac{y^2}{4}=1, \) find the eccentricity.
▶️ Answer / Explanation
Here \( a^2=9,\ b^2=4 \).
\( c^2 = a^2 + b^2 = 9 + 4 = 13 \Rightarrow c = \sqrt{13} \)
\( e = \dfrac{c}{a} = \dfrac{\sqrt{13}}{3} \)
Answer: \( e = \dfrac{\sqrt{13}}{3} \)
Example
Find the asymptotes of hyperbola \( \dfrac{x^2}{25}-\dfrac{y^2}{16}=1. \)
▶️ Answer / Explanation
\( a^2=25 \Rightarrow a=5,\quad b^2=16 \Rightarrow b=4 \)
Asymptotes:
\( y = \pm\dfrac{b}{a}x = \pm\dfrac{4}{5}x \)
Answer: \( y = \frac{4}{5}x,\quad y = -\frac{4}{5}x \)
Example
For hyperbola \( \dfrac{y^2}{36}-\dfrac{x^2}{20}=1, \) find its center, vertices, and foci.
▶️ Answer / Explanation
This is a vertical hyperbola.
\( a^2 = 36 \Rightarrow a = 6 \)
\( b^2 = 20 \Rightarrow b = 2\sqrt{5} \)
\( c^2 = a^2 + b^2 = 36 + 20 = 56 \Rightarrow c = \sqrt{56} = 2\sqrt{14} \)
Center: \( (0,0) \)
Vertices: \( (0,\pm a) = (0,\pm 6) \)
Foci: \( (0,\pm c) = (0,\pm 2\sqrt{14}) \)
Hyperbola: Parametric Form
Parametric coordinates make hyperbola problems involving tangent, normal, chord lengths, and locus extremely easy and are used frequently in JEE.
Standard hyperbola:
\( \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 \)
1. Parametric Coordinates of Hyperbola
The most useful parametric form (eccentric-parameter form) is:
\( (x,y) = (a\sec\theta,\ b\tan\theta) \)
Here \( \theta \) is called the hyperbolic parameter.
Verification:
\( \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} = \sec^2\theta – \tan^2\theta = 1 \)
Thus every point of the hyperbola is represented uniquely by some \( \theta \).
2. Parametric Form for Vertical Hyperbola
If hyperbola is:
\( \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 \)
Then parametric form becomes:
\( (x,y) = (b\tan\theta,\ a\sec\theta) \)
3. Alternative Parametric Form (Hyperbolic Functions)
This is less used in JEE, but good for theoretical problems:
\( x = a\cosh t,\ y = b\sinh t \)
Here \( t \) is the hyperbolic angle. Identity:
\( \cosh^2 t – \sinh^2 t = 1 \)
4. Conjugate Hyperbola Parametric Form
Conjugate hyperbola:
\( \dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1 \)
Parametric coordinates:
\( (x,y)=(a\tan\theta,\ b\sec\theta) \)
5. Key Properties Using Parametric Form
- Chord length between points \(\theta_1\) and \(\theta_2\):
\( PQ=\sqrt{a^2(\sec\theta_1 – \sec\theta_2)^2 + b^2(\tan\theta_1 – \tan\theta_2)^2} \) - Tangent at \((a\sec\theta,\ b\tan\theta)\):
\( \dfrac{x\sec\theta}{a} – \dfrac{y\tan\theta}{b} = 1 \) - Normal at parametric point:
\( ax\cos\theta + by\sin\theta = a^2 – b^2 \) - Distance of parametric point from center = \( \sqrt{a^2\sec^2\theta + b^2\tan^2\theta} \)
6. Parametric Form & Asymptotes
As \( \theta \to \dfrac{\pi}{2} \): \( \sec\theta \to \infty,\ \tan\theta \to \infty \)
Then the point \((a\sec\theta,\ b\tan\theta)\) approaches the lines:
\( y=\pm\dfrac{b}{a}x \)
Hence parametric form naturally shows asymptotic behavior.
Example
Find the parametric point corresponding to \( \theta = \dfrac{\pi}{4} \) for hyperbola \( \dfrac{x^2}{9}-\dfrac{y^2}{4}=1. \)
▶️ Answer / Explanation
Here \( a=3,\ b=2 \).
Parametric form:
\( x = a\sec\theta = 3\sec\dfrac{\pi}{4} = 3\sqrt{2} \)
\( y = b\tan\theta = 2\tan\dfrac{\pi}{4} = 2 \)
Answer: \( (3\sqrt{2},\ 2) \)
Example
Find the coordinates of points where the hyperbola \( \dfrac{x^2}{16}-\dfrac{y^2}{9}=1 \) intersects the line \( y = 2x \) using parametric form.
▶️ Answer / Explanation
Parametric form for hyperbola:
\( x = 4\sec\theta,\ y = 3\tan\theta \)
Substitute in \( y = 2x \):
\( 3\tan\theta = 2(4\sec\theta) \)
\( 3\tan\theta = 8\sec\theta \)
Divide by \(\cos\theta\):
\( 3\sin\theta = 8 \)
Impossible since \( |\sin\theta| \le 1 \). Thus the line does NOT cut hyperbola.
Answer: No real intersection points.
Example
Find the equation of the chord joining parametric points \( \theta = \alpha \) and \( \theta = \beta \) for hyperbola \( \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1. \)
▶️ Answer / Explanation
Parametric points:
\( P(a\sec\alpha,\ b\tan\alpha) \)
\( Q(a\sec\beta,\ b\tan\beta) \)
Equation of line joining P and Q using 2-point form:
\( \dfrac{y – b\tan\alpha}{b\tan\beta – b\tan\alpha} = \dfrac{x – a\sec\alpha}{a\sec\beta – a\sec\alpha} \)
Simplify:
\( \dfrac{y – b\tan\alpha}{b(\tan\beta – \tan\alpha)} = \dfrac{x – a\sec\alpha}{a(\sec\beta – \sec\alpha)} \)
Cross multiplying:
\( a(y – b\tan\alpha)(\sec\beta – \sec\alpha) = b(x – a\sec\alpha)(\tan\beta – \tan\alpha) \)
This is the required chord equation.
Final Answer:
\( a(y – b\tan\alpha)(\sec\beta – \sec\alpha) = b(x – a\sec\alpha)(\tan\beta – \tan\alpha) \)
Notes and Study Materials
- Concepts of Hyperbola
- Hyperbola Master File
- Hyperbola Revision Notes
- Hyperbola Formulae
- Hyperbola Reference Book
- Hyperbola Past Many Years Questions and Answer
Examples and Exercise
IIT JEE (Main) Mathematics ,”Hyperbola” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian
About this unit
Hyperbola
IITian Academy Notes for IIT JEE (Main) Mathematics – Hyperbola
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IIT JEE (Main) Mathematics, Hyperbola Solved Examples and Practice Papers.
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S. L. Loney IIT JEE (Main) Mathematics
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Hall & Knight IIT JEE (Main) Mathematics
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