How to Solve Hyperbola (IIT JEE Main + Advanced) – Complete Guide

How to Solve Hyperbola (IIT JEE Main + Advanced) – Complete Guide

Introduction

"Mastering hyperbolas unlocks 8–12 marks in IIT JEE—enough to push you from a 90 to a 100+ percentile. Whether it’s finding asymptotes in 30 seconds or deriving tangents under time pressure, this guide gives you the exact steps to solve any hyperbola problem on exam day."

What You Need To Know First

1. Standard equations of conic sections (parabola, ellipse, hyperbola).
2. Basic coordinate geometry (distance formula, slope of a line).
3. Differentiation (for tangent/normal problems).

Key Vocabulary

Formulas To Know

1. Standard Equation of Hyperbola

[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{(Transverse axis along x-axis)} ] - (a) = distance from center to vertex. - (b) = distance from center to co-vertex. - MEMORISE THIS.

2. Asymptotes of Hyperbola

[ y = \pm \frac{b}{a}x \quad \text{(for (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1))} ] - MEMORISE THIS.

3. Rectangular Hyperbola

[ xy = c^2 \quad \text{(Asymptotes: } x=0, y=0\text{)} ] - MEMORISE THIS.

4. Latus Rectum Length

[ \text{Length} = \frac{2b^2}{a} ] - MEMORISE THIS.

5. Tangent to Hyperbola (Point Form)

[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \quad \text{(for (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1) at point ((x_1, y_1)))} ] - MEMORISE THIS.

6. Tangent to Hyperbola (Slope Form)

[ y = mx \pm \sqrt{a^2 m^2 - b^2} \quad \text{(for (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1))} ] - MEMORISE THIS.

7. Normal to Hyperbola (Point Form)

[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 \quad \text{(for (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1) at ((x_1, y_1)))} ] - MEMORISE THIS.

8. Eccentricity (e)

[ e = \sqrt{1 + \frac{b^2}{a^2}} ] - MEMORISE THIS.

Step-by-Step Method

Step 1: Identify the Hyperbola Type

• If equation is (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1) → Horizontal hyperbola.
• If equation is (\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1) → Vertical hyperbola.
• If equation is (xy = c^2) → Rectangular hyperbola.

Step 2: Find Asymptotes

• For (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1) → Asymptotes: (y = \pm \frac{b}{a}x).
• For (\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1) → Asymptotes: (y = \pm \frac{a}{b}x).
• For (xy = c^2) → Asymptotes: (x=0, y=0).

Step 3: Find Latus Rectum

• Length = (\frac{2b^2}{a}) (for (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1)).
• Endpoints: ((\pm ae, \pm \frac{b^2}{a})).

Step 4: Find Tangent at a Point

• Use point form: (\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1).
• Or slope form: (y = mx \pm \sqrt{a^2 m^2 - b^2}).

Step 5: Find Normal at a Point

• Use point form: (\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2).

Step 6: Check for Rectangular Hyperbola

• If (a = b) → Rectangular hyperbola.
• Asymptotes are perpendicular.

Worked Examples

Example 1 – Basic (Asymptotes & Latus Rectum)

Problem: Find the asymptotes and latus rectum of (\frac{x^2}{16} - \frac{y^2}{9} = 1).

Solution: 1. Identify (a) and (b):
- (a^2 = 16 \Rightarrow a = 4)
- (b^2 = 9 \Rightarrow b = 3)

1. Asymptotes:

2. (y = \pm \frac{b}{a}x = \pm \frac{3}{4}x)

3. Latus Rectum:

4. Length = (\frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2})

Answer: - Asymptotes: (y = \pm \frac{3}{4}x) - Latus Rectum: (\frac{9}{2})

What we did and why: - We used the standard formulas for asymptotes and latus rectum. No tricks—just plug and play.

Example 2 – Medium (Tangent & Normal)

Problem: Find the equation of the tangent and normal to (\frac{x^2}{25} - \frac{y^2}{16} = 1) at ((5\sqrt{2}, 4)).

Solution: 1. Verify point lies on hyperbola:
- (\frac{(5\sqrt{2})^2}{25} - \frac{4^2}{16} = \frac{50}{25} - \frac{16}{16} = 2 - 1 = 1) ✔️

1. Tangent (Point Form):
2. (\frac{xx_1}{25} - \frac{yy_1}{16} = 1)
3. (\frac{x(5\sqrt{2})}{25} - \frac{y(4)}{16} = 1)

4. (\frac{\sqrt{2}x}{5} - \frac{y}{4} = 1)

5. Normal (Point Form):

6. (\frac{25x}{5\sqrt{2}} + \frac{16y}{4} = 25 + 16)
7. (\frac{5x}{\sqrt{2}} + 4y = 41)

Answer: - Tangent: (\frac{\sqrt{2}x}{5} - \frac{y}{4} = 1) - Normal: (\frac{5x}{\sqrt{2}} + 4y = 41)

What we did and why: - We used the point form of tangent and normal. Always verify the point lies on the hyperbola first!

Example 3 – Exam-Style (Disguised Problem)

Problem: A hyperbola has asymptotes (y = \pm \frac{3}{2}x) and passes through ((4, 3)). Find its equation.

Solution: 1. From asymptotes, find (\frac{b}{a}):
- (\frac{b}{a} = \frac{3}{2} \Rightarrow b = \frac{3}{2}a)

1. Standard equation:
2. (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1)

3. Substitute (b = \frac{3}{2}a):
   (\frac{x^2}{a^2} - \frac{y^2}{(\frac{9}{4}a^2)} = 1)

4. Use point ((4, 3)):

5. (\frac{16}{a^2} - \frac{9}{\frac{9}{4}a^2} = 1)
6. (\frac{16}{a^2} - \frac{4}{a^2} = 1)

7. (\frac{12}{a^2} = 1 \Rightarrow a^2 = 12)

8. Find (b^2):

9. (b^2 = \frac{9}{4}a^2 = \frac{9}{4} \times 12 = 27)

10. Final equation:

11. (\frac{x^2}{12} - \frac{y^2}{27} = 1)

Answer: (\frac{x^2}{12} - \frac{y^2}{27} = 1)

What we did and why: - We used the asymptotes to relate (a) and (b), then substituted a point to find exact values. This is a common IIT JEE trick—don’t assume (a) and (b) are given directly!

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—hyperbolas are all about three things: asymptotes, tangents, and the latus rectum. Memorise these formulas cold:

1. Standard equation: (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1).
2. Asymptotes: (y = \pm \frac{b}{a}x).
3. Tangent at ((x_1, y_1)): (\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1).
4. Latus rectum: (\frac{2b^2}{a}).

If the problem gives asymptotes, relate (b/a). If it asks for tangents, use point form or slope form. For rectangular hyperbolas ((xy = c^2)), remember asymptotes are the axes. Double-check your signs—hyperbolas have minus signs, ellipses have plus. You’ve got this!