Chapter 10 — CONIC SECTIONS

Concept & Theory

The standard equation of a hyperbola with transverse axis along the x-axis is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

• Centre: \((0,0)\)
• Vertices: \((\pm a,0)\)
• Foci: \((\pm c,0)\)
• Relation: \(c^2 = a^2 + b^2\)
• Eccentricity: \(e = \dfrac{c}{a}\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)

Solution Roadmap

1. Compare with standard form
2. Extract \(a^2\), \(b^2\)
3. Compute \(c\) using \(c^2 = a^2 + b^2\)
4. Find foci and vertices
5. Compute eccentricity
6. Find latus rectum length

Solution

Given: \[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]

Comparing with standard form: \[ a^2 = 16,\quad b^2 = 9 \] \[ a = 4,\quad b = 3 \]

Using \(c^2 = a^2 + b^2\): \[ c = \sqrt{16 + 9} = 5 \]

Foci: \((\pm 5, 0)\)

Vertices: \((\pm 4, 0)\)

Eccentricity: \[ \begin{aligned} e &= \frac{c}{a} \\&= \frac{5}{4} \end{aligned} \]

Latus rectum length: \[ \begin{aligned} \frac{2b^2}{a} &= \frac{2 \times 9}{4} \\&= \frac{9}{2} \end{aligned} \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Direct formula-based question (very high scoring, zero conceptual ambiguity)
• JEE Main: Frequently asked as objective-type identification or parameter extraction
• JEE Advanced: Used as a base to build multi-concept problems (asymptotes, parametric form, etc.)
• Key Skill: Fast recognition of standard form → saves significant time

Concept & Theory

The standard equation of a hyperbola with transverse axis along the y-axis is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Centre: \((0,0)\)
• Vertices: \((0,\pm a)\)
• Foci: \((0,\pm c)\)
• Relation: \(c^2 = a^2 + b^2\)
• Eccentricity: \(e = \dfrac{c}{a}\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)

Solution Roadmap

1. Identify orientation (y² positive → vertical hyperbola)
2. Extract \(a^2\), \(b^2\)
3. Compute \(c\)
4. Write foci and vertices
5. Find eccentricity
6. Compute latus rectum

Solution

Given: \[ \frac{y^2}{9} - \frac{x^2}{27} = 1 \]

Comparing with standard form: \[ a^2 = 9,\quad b^2 = 27 \] \[ a = 3,\quad b = 3\sqrt{3} \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} c &= \sqrt{9 + 27} \\&= \sqrt{36} \\&= 6 \end{aligned} \]

Foci: \((0, \pm 6)\)

Vertices: \((0, \pm 3)\)

Eccentricity: \[ \begin{aligned} e &= \frac{c}{a} \\&= \frac{6}{3} \\&= 2 \end{aligned} \]

Latus rectum length: \[ \begin{aligned} \frac{2b^2}{a} &= \frac{2 \times 27}{3} \\&= 18 \end{aligned} \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Tests identification of vertical hyperbola (very common pattern)
• JEE Main: Direct parameter extraction and orientation-based MCQs
• JEE Advanced: Often extended to asymptotes, conjugate axis, or transformation problems
• Key Insight: Sign of terms immediately reveals orientation → saves time

Concept & Theory

A hyperbola with transverse axis along the y-axis is written as: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Centre: \((0,0)\)
• Vertices: \((0,\pm a)\)
• Foci: \((0,\pm c)\)
• Relation: \(c^2 = a^2 + b^2\)
• Eccentricity: \(e = \dfrac{c}{a}\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)

Solution Roadmap

1. Convert equation into standard form
2. Identify orientation (y² positive → vertical hyperbola)
3. Extract \(a\), \(b\)
4. Compute \(c\)
5. Find foci, vertices
6. Calculate eccentricity and latus rectum

Solution

Given: \[ 9y^2 - 4x^2 = 36 \]

Divide both sides by 36: \[ \begin{aligned} \frac{9y^2}{36} - \frac{4x^2}{36} &= 1\\ \Rightarrow \frac{y^2}{4} - \frac{x^2}{9} &= 1 \end{aligned} \]

Comparing with standard form: \[ a^2 = 4,\quad b^2 = 9 \] \[ a = 2,\quad b = 3 \]

Using \(c^2 = a^2 + b^2\): \[ c = \sqrt{4 + 9} = \sqrt{13} \]

Foci: \((0, \pm \sqrt{13})\)

Vertices: \((0, \pm 2)\)

Eccentricity: \[ e = \frac{\sqrt{13}}{2} \]

Latus rectum length: \[ \frac{2b^2}{a} = \frac{2 \times 9}{2} = 9 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Tests ability to convert general equation into standard form
• JEE Main: Frequently asked in normalized form identification problems
• JEE Advanced: Used in problems involving asymptotes and coordinate transformations
• Key Skill: Fast normalization (divide by constant) is critical for accuracy

Concept & Theory

A hyperbola with transverse axis along the x-axis is written as: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

• Centre: \((0,0)\)
• Vertices: \((\pm a,0)\)
• Foci: \((\pm c,0)\)
• Relation: \(c^2 = a^2 + b^2\)
• Eccentricity: \(e = \dfrac{c}{a}\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)

Solution Roadmap

1. Convert to standard form
2. Identify orientation (x² positive → horizontal hyperbola)
3. Extract \(a\), \(b\)
4. Compute \(c\)
5. Find foci and vertices
6. Compute eccentricity and latus rectum

Solution

Given: \[ 16x^2 - 9y^2 = 576 \]

Divide both sides by 576: \[ \frac{16x^2}{576} - \frac{9y^2}{576} = 1 \Rightarrow \frac{x^2}{36} - \frac{y^2}{64} = 1 \]

Comparing with standard form: \[ a^2 = 36,\quad b^2 = 64 \] \[ a = 6,\quad b = 8 \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} c &= \sqrt{36 + 64} \\&= \sqrt{100} \\&= 10 \end{aligned} \]

Foci: \((\pm 10, 0)\)

Vertices: \((\pm 6, 0)\)

Eccentricity: \[ \begin{aligned} e &= \frac{c}{a} \\&= \frac{10}{6} \\&= \frac{5}{3} \end{aligned} \]

Latus rectum length: \[ \begin{aligned} \frac{2b^2}{a} &= \frac{2 \times 64}{6} \\&= \frac{64}{3} \end{aligned} \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Classic normalization + parameter extraction problem
• JEE Main: Direct formula substitution with simplification
• JEE Advanced: Often linked with asymptotes and rectangular hyperbola concepts
• Key Skill: Recognizing structure quickly avoids algebraic mistakes

Concept & Theory

A hyperbola with vertical transverse axis is expressed as: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Centre: \((0,0)\)
• Vertices: \((0,\pm a)\)
• Foci: \((0,\pm c)\)
• Relation: \(c^2 = a^2 + b^2\)
• Eccentricity: \(e = \dfrac{c}{a}\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)

Solution Roadmap

1. Convert to standard form
2. Identify orientation (y² positive → vertical)
3. Extract \(a^2, b^2\)
4. Compute \(c\) carefully (fraction handling)
5. Find foci, vertices
6. Simplify eccentricity properly

Solution

Given: \[ 5y^2 - 9x^2 = 36 \]

Divide both sides by 36: \[ \begin{aligned} \frac{5y^2}{36} - \frac{9x^2}{36} &= 1\\ \Rightarrow \frac{y^2}{36/5} - \frac{x^2}{4} &= 1 \end{aligned} \]

Comparing with standard form: \[ a^2 = \frac{36}{5}, \quad b^2 = 4 \] \[ a = \frac{6}{\sqrt{5}}, \quad b = 2 \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} c &= \sqrt{\frac{36}{5} + 4}\\ &= \sqrt{\frac{36 + 20}{5}}\\ &= \sqrt{\frac{56}{5}}\\ &= \frac{2\sqrt{14}}{\sqrt{5}} \end{aligned} \]

Foci: \(\left(0, \pm \frac{2\sqrt{14}}{\sqrt{5}}\right)\)

Vertices: \(\left(0, \pm \frac{6}{\sqrt{5}}\right)\)

Eccentricity: \[ \begin{aligned} e &= \frac{c}{a}\\ &= \frac{\frac{2\sqrt{14}}{\sqrt{5}}}{\frac{6}{\sqrt{5}}}\\ &= \frac{2\sqrt{14}}{6}\\ &= \frac{\sqrt{14}}{3} \end{aligned} \]

Latus rectum length: \[ \begin{aligned} \frac{2b^2}{a} &= \frac{2 \times 4}{6/\sqrt{5}}\\ &= \frac{8\sqrt{5}}{6}\\ &= \frac{4\sqrt{5}}{3} \end{aligned} \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Tests handling of fractional \(a^2\) → common mistake zone
• JEE Main: Focus on simplification accuracy (irrational expressions)
• JEE Advanced: Often embedded in multi-step algebra-heavy problems
• Key Skill: Avoid early approximation → keep radicals exact till final

Concept & Theory

A hyperbola with transverse axis along the y-axis has the standard form: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Centre: \((0,0)\)
• Vertices: \((0,\pm a)\)
• Foci: \((0,\pm c)\)
• Relation: \(c^2 = a^2 + b^2\)
• Eccentricity: \(e = \dfrac{c}{a} > 1\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)

Solution Roadmap

1. Reduce equation to standard form
2. Identify orientation (y² positive → vertical hyperbola)
3. Extract \(a^2\), \(b^2\)
4. Compute \(c\)
5. Write foci and vertices
6. Compute eccentricity and latus rectum (keep exact form)

Solution

Given: \[ 49y^2 - 16x^2 = 784 \]

Divide both sides by 784: \[ \begin{aligned} \frac{49y^2}{784} - \frac{16x^2}{784} &= 1\\ \Rightarrow \frac{y^2}{16} - \frac{x^2}{49} &= 1 \end{aligned} \]

Comparing with standard form: \[ a^2 = 16,\quad b^2 = 49 \] \[ a = 4,\quad b = 7 \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} c &= \sqrt{16 + 49} \\&= \sqrt{65} \end{aligned} \]

Foci: \((0, \pm \sqrt{65})\)

Vertices: \((0, \pm 4)\)

Eccentricity: \[ \begin{aligned} e &= \frac{c}{a} \\&= \frac{\sqrt{65}}{4} \end{aligned} \]

Latus rectum length: \[ \begin{aligned} \frac{2b^2}{a} &= \frac{2 \times 49}{4} \\&= \frac{98}{4} \\&= \frac{49}{2} \end{aligned} \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Direct standard form conversion + parameter extraction (very high scoring)
• JEE Main: Focus on identifying correct orientation and avoiding arithmetic slips
• JEE Advanced: Acts as base for asymptotes, director circle, and parametric questions
• Common Mistake: Writing decimal \(24.5\) instead of exact form \(\frac{49}{2}\)
• Key Skill: Always keep answers in exact fractional/radical form unless asked otherwise

Concept & Theory

For a hyperbola centered at the origin with transverse axis along the x-axis: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

• Vertices: \((\pm a, 0)\)
• Foci: \((\pm c, 0)\)
• Relation: \(c^2 = a^2 + b^2\)

If both vertices and foci lie on the x-axis, the hyperbola is horizontal.

Solution Roadmap

1. Identify orientation from given points
2. Extract \(a\) and \(c\)
3. Use relation \(c^2 = a^2 + b^2\) to find \(b^2\)
4. Substitute into standard equation

Solution

Given: Vertices \((\pm 2, 0)\), foci \((\pm 3, 0)\)

Since both lie on the x-axis, the hyperbola has a horizontal transverse axis.

\[ a = 2,\quad c = 3 \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} b^2 &= c^2 - a^2 \\&= 9 - 4 \\&= 5 \end{aligned} \]

Substituting into standard form:

Equation of hyperbola: \[ \frac{x^2}{4} - \frac{y^2}{5} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Direct concept application of identifying \(a\), \(c\)
• JEE Main: Frequently appears as parameter-based MCQ
• JEE Advanced: Used as base for asymptotes or parametric extensions
• Key Insight: Orientation is immediately determined by alignment of vertices and foci
• Common Mistake: Using wrong formula \(b^2 = a^2 - c^2\) (valid for ellipse, not hyperbola)

Concept & Theory

For a hyperbola centered at the origin with transverse axis along the y-axis: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Vertices: \((0, \pm a)\)
• Foci: \((0, \pm c)\)
• Relation: \(c^2 = a^2 + b^2\)

If vertices and foci lie on the y-axis, the hyperbola is vertical.

Solution Roadmap

1. Determine orientation from given coordinates
2. Extract \(a\) and \(c\)
3. Use \(c^2 = a^2 + b^2\) to find \(b^2\)
4. Substitute into standard equation

Solution

Given: Vertices \((0, \pm 5)\), foci \((0, \pm 8)\)

Since both lie on the y-axis, the hyperbola has a vertical transverse axis.

\[ a = 5,\quad c = 8 \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} b^2 &= c^2 - a^2 \\&= 64 - 25 \\&= 39 \end{aligned} \]

Geometric Insight (Graph)

Equation of hyperbola: \[ \frac{y^2}{25} - \frac{x^2}{39} = 1 \]

Exam Significance

• Board Exams: Straightforward parameter identification and substitution
• JEE Main: Orientation-based question (horizontal vs vertical recognition)
• JEE Advanced: Can extend to asymptotes or transformation problems
• Key Insight: Position of vertices immediately determines equation form
• Common Mistake: Writing \(b^2 = a^2 - c^2\) (ellipse formula confusion)

Concept & Theory

For a hyperbola centered at the origin with transverse axis along the y-axis: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Vertices: \((0, \pm a)\)
• Foci: \((0, \pm c)\)
• Relation: \(c^2 = a^2 + b^2\)

Alignment of vertices and foci along the y-axis confirms a vertical hyperbola.

Solution Roadmap

1. Identify orientation from coordinates
2. Extract \(a\) and \(c\)
3. Compute \(b^2\) using \(c^2 = a^2 + b^2\)
4. Write standard equation

Solution

Given: Vertices \((0, \pm 3)\), foci \((0, \pm 5)\)

Since both lie on the y-axis, the hyperbola is vertical.

\[ a = 3,\quad c = 5 \]

Using \(c^2 = a^2 + b^2\): \[ b^2 = 25 - 9 = 16 \]

Equation of hyperbola: \[ \frac{y^2}{9} - \frac{x^2}{16} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Straightforward substitution-based question
• JEE Main: Frequently appears in parameter-identification MCQs
• JEE Advanced: May be extended to asymptotes or eccentricity-based problems
• Key Insight: Perfect square values (9,16,25) make this a high-speed question
• Common Mistake: Mixing horizontal and vertical standard forms

JEE Quick Trigger

If vertices and foci are \((0, \pm a)\), \((0, \pm c)\) → directly write: \[ \frac{y^2}{a^2} - \frac{x^2}{c^2 - a^2} = 1 \]

Concept & Theory

For a hyperbola centered at the origin with transverse axis along the x-axis: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

• Length of transverse axis = \(2a\)
• Foci: \((\pm c, 0)\)
• Relation: \(c^2 = a^2 + b^2\)

If foci lie on the x-axis, the hyperbola is horizontal.

Solution Roadmap

1. Convert transverse axis length → find \(a\)
2. Extract \(c\) from foci
3. Compute \(b^2\)
4. Write standard equation

Solution

Given: Foci \((\pm 5, 0)\), transverse axis length \(8\)

\[ 2a = 8 \Rightarrow a = 4,\quad c = 5 \]

Since foci lie on x-axis, the hyperbola is horizontal.

Using \(c^2 = a^2 + b^2\): \[ b^2 = 25 - 16 = 9 \]

Equation of hyperbola: \[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Tests understanding of transverse axis length concept
• JEE Main: Very common pattern combining geometry + formula
• JEE Advanced: May extend to eccentricity or asymptote-based questions
• Key Insight: Always convert “length” → parameter before using formulas
• Common Mistake: Taking \(a = 8\) instead of \(a = 4\)

JEE Quick Trigger

If foci are \((\pm c,0)\) and transverse axis is given → directly: \[ a = \frac{\text{length}}{2}, \quad b^2 = c^2 - a^2 \]

Concept & Theory

For a hyperbola centered at the origin with transverse axis along the y-axis: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Foci: \((0, \pm c)\)
• Length of conjugate axis = \(2b\)
• Relation: \(c^2 = a^2 + b^2\)

If foci lie on the y-axis, the hyperbola is vertical.

Solution Roadmap

1. Convert conjugate axis length → find \(b\)
2. Extract \(c\) from foci
3. Use \(c^2 = a^2 + b^2\) to find \(a^2\)
4. Write standard equation

Solution

Given: Foci \((0, \pm 13)\), conjugate axis length \(24\)

\[ \begin{aligned} 2b &= 24 \\ \Rightarrow b &= 12,\\ b^2 &= 144 \end{aligned} \]

Since foci lie on the y-axis, the hyperbola is vertical.

\[ c = 13 \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} a^2 &= c^2 - b^2 \\&= 169 - 144 \\&= 25 \end{aligned} \]

\[ a = 5 \]

Equation of hyperbola: \[ \frac{y^2}{25} - \frac{x^2}{144} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Tests understanding of conjugate axis (commonly confused term)
• JEE Main: Multi-step parameter extraction (length → parameter → equation)
• JEE Advanced: Can be extended to asymptotes or rectangular hyperbola checks
• Key Insight: Use \(a^2 = c^2 - b^2\) when \(b\) is given directly
• Common Mistake: Confusing transverse axis with conjugate axis

JEE Quick Trigger

If conjugate axis is given: \[ b = \frac{\text{length}}{2}, \quad a^2 = c^2 - b^2 \]

Concept & Theory

For a hyperbola with transverse axis along the x-axis: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

• Foci: \((\pm c, 0)\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)
• Relation: \(c^2 = a^2 + b^2\)

This type of problem requires forming and solving an equation in \(a\).

Solution Roadmap

1. Extract \(c\) from foci
2. Use latus rectum formula to express \(b^2\) in terms of \(a\)
3. Substitute into \(c^2 = a^2 + b^2\)
4. Solve quadratic for \(a\)
5. Find \(b^2\) and write equation

Solution

Given: Foci \((\pm 3\sqrt{5}, 0)\), latus rectum length \(8\)

\[ \begin{aligned} c &= 3\sqrt{5} \\\Rightarrow c^2 &= 45 \end{aligned} \]

Using latus rectum formula: \[ \begin{aligned} \frac{2b^2}{a} &= 8 \\ \Rightarrow b^2 &= 4a \end{aligned} \]

Substitute into \(c^2 = a^2 + b^2\): \[ \begin{aligned} 45 &= a^2 + 4a\\ \Rightarrow a^2 + 4a - 45 &= 0 \end{aligned} \]

Solving: \[ \begin{aligned} a &= \frac{-4 \pm \sqrt{16 + 180}}{2}\\ &= \frac{-4 \pm \sqrt{196}}{2}\\ &= \frac{-4 \pm 14}{2} \end{aligned} \]

Valid solution: \[ a = 5 \quad (\text{reject negative value}) \]

\[ b^2 = 4a = 20 \]

Equation of hyperbola: \[ \frac{x^2}{25} - \frac{y^2}{20} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Higher-level problem involving equation formation
• JEE Main: Very common mixed-parameter question (focus + latus rectum)
• JEE Advanced: Classic quadratic-in-parameter setup
• Key Insight: Convert everything into one variable (usually \(a\))
• Common Mistake: Forgetting to reject negative root of \(a\)

JEE Quick Trigger

If latus rectum is given: \[ b^2 = \frac{(\text{latus rectum}) \cdot a}{2} \Rightarrow \text{plug into } c^2 = a^2 + b^2 \]

Concept & Theory

For a hyperbola with transverse axis along the x-axis: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

• Foci: \((\pm c, 0)\)
• Latus rectum length: \(\dfrac{2b^2}{a}\)
• Relation: \(c^2 = a^2 + b^2\)

These problems reduce to forming a quadratic equation in \(a\).

Solution Roadmap

1. Extract \(c\) from foci
2. Use latus rectum to express \(b^2\) in terms of \(a\)
3. Substitute into \(c^2 = a^2 + b^2\)
4. Solve quadratic and select valid root
5. Form equation

Solution

Given: Foci \((\pm 4, 0)\), latus rectum length \(12\)

\[ c = 4 \Rightarrow c^2 = 16 \]

Using latus rectum formula: \[ \frac{2b^2}{a} = 12 \Rightarrow b^2 = 6a \]

Substitute into \(c^2 = a^2 + b^2\): \[ 16 = a^2 + 6a \Rightarrow a^2 + 6a - 16 = 0 \]

Solving: \[ a = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm 10}{2} \]

Valid solution: \[ a = 2 \quad (\text{reject negative value}) \]

\[ b^2 = 6a = 12 \]

Equation of hyperbola: \[ \frac{x^2}{4} - \frac{y^2}{12} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Tests quadratic formation using geometric parameters
• JEE Main: Common mixed-condition problem (focus + latus rectum)
• JEE Advanced: Prototype for parameter-solving questions
• Key Insight: Express everything in terms of one variable before solving
• Common Mistake: Not simplifying \(b^2 = 6a\) correctly or picking negative root

JEE Quick Trigger

If focus + latus rectum given: \[ b^2 = \frac{(\text{latus rectum}) \cdot a}{2} \Rightarrow \text{plug into } c^2 = a^2 + b^2 \]

Concept & Theory

For a hyperbola with transverse axis along the x-axis: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

• Vertices: \((\pm a, 0)\)
• Eccentricity: \(e = \dfrac{c}{a}\)
• Relation: \(c^2 = a^2 + b^2\)

Eccentricity-based problems are solved by first finding \(c\), then \(b^2\).

Solution Roadmap

1. Extract \(a\) from vertices
2. Use \(e = \frac{c}{a}\) to find \(c\)
3. Use \(c^2 = a^2 + b^2\) to compute \(b^2\)
4. Write equation (prefer clean denominator form)

Solution

Given: Vertices \((\pm 7, 0)\), \(e = \frac{4}{3}\)

Since vertices lie on x-axis, the hyperbola is horizontal.

\[ a = 7 \]

Using \(e = \frac{c}{a}\): \[ \begin{aligned} c &= e \cdot a \\&= \frac{4}{3} \times 7 \\&= \frac{28}{3} \end{aligned} \]

Using \(c^2 = a^2 + b^2\): \[ \begin{aligned} b^2 &= c^2 - a^2\\ &= \frac{784}{9} - 49\\ &= \frac{784 - 441}{9}\\ &= \frac{343}{9} \end{aligned} \]

Equation of hyperbola:

Standard form: \[ \frac{x^2}{49} - \frac{y^2}{343/9} = 1 \]

Preferred simplified form: \[ \frac{x^2}{49} - \frac{9y^2}{343} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: Tests eccentricity application with fraction handling
• JEE Main: Common question involving \(e = \frac{c}{a}\)
• JEE Advanced: Often combined with asymptotes or parametric forms
• Key Insight: Always compute \(c\) first in eccentricity problems
• Common Mistake: Not converting mixed fractions properly in \(b^2\)

JEE Quick Trigger

If eccentricity is given: \[ c = e \cdot a,\quad b^2 = a^2(e^2 - 1) \] (direct shortcut, avoids two-step calculation)

Concept & Theory

For a hyperbola with transverse axis along the y-axis: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

• Foci: \((0, \pm c)\)
• Relation: \(c^2 = a^2 + b^2\)
• Any point on the hyperbola satisfies its equation

When a point is given, substitute it directly into the standard equation.

Solution Roadmap

1. Identify orientation from foci
2. Use \(c^2 = a^2 + b^2\)
3. Substitute given point into equation
4. Form equation in one variable
5. Solve and back-substitute

Solution

Given: Foci \((0, \pm \sqrt{10})\), point \((2,3)\)

Since foci lie on the y-axis, the hyperbola is vertical.

\[ c = \sqrt{10} \Rightarrow c^2 = 10 \]

Using \(c^2 = a^2 + b^2\): \[ a^2 + b^2 = 10 \tag{1} \]

Substitute point \((2,3)\) into: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

\[ \frac{9}{a^2} - \frac{4}{b^2} = 1 \tag{2} \]

Let \(b^2 = t\), then from (1): \[ a^2 = 10 - t \]

Substitute into (2): \[ \frac{9}{10 - t} - \frac{4}{t} = 1 \]

Multiply throughout by \(t(10 - t)\): \[ 9t - 4(10 - t) = t(10 - t) \]

\[ \begin{aligned} 9t - 40 + 4t &= 10t - t^2\\ \Rightarrow 13t - 40 &= 10t - t^2 \end{aligned} \]

\[ t^2 + 3t - 40 = 0 \]

\[ (t + 8)(t - 5) = 0 \Rightarrow t = 5 \quad (\text{reject negative}) \]

\[ b^2 = 5,\quad a^2 = 10 - 5 = 5 \]

Equation of hyperbola: \[ \frac{y^2}{5} - \frac{x^2}{5} = 1 \]

Geometric Insight (Graph)

Exam Significance

• Board Exams: High-level problem combining point + focus condition
• JEE Main: Classic substitution-based parameter solving
• JEE Advanced: Standard model for multi-variable reduction problems
• Key Insight: Always reduce to one variable before solving
• Common Mistake: Substituting point incorrectly or mishandling fractions

JEE Quick Trigger

If a point is given: convert everything to one variable (usually \(b^2\)), then solve systematically.