Chapter 10 — CONIC SECTIONS

Theory

Standard form of ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If \(a>b\): major axis along x-axis
• Foci: \((\pm c,0)\), where \(c=\sqrt{a^2-b^2}\)
• Vertices: \((\pm a,0)\)
• Major axis length: \(2a\)
• Minor axis length: \(2b\)
• Eccentricity: \(e=\frac{c}{a}\)
• Latus rectum: \(\frac{2b^2}{a}\)

Solution Roadmap

• Compare with standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{a^2-b^2}\)
• Find required parameters

Solution

Comparing with standard form:

\[ a^2=36,\quad b^2=16 \]

Since \(a^2>b^2\), major axis lies along x-axis.

\[ a=6,\quad b=4 \] \[ c=\sqrt{36-16}=\sqrt{20} \]

Foci:

\[ (\pm \sqrt{20},0) \]

Vertices:

\[ (\pm 6,0) \]

Major axis length:

\[ 2a=12 \]

Minor axis length:

\[ 2b=8 \]

Eccentricity:

\[ e=\frac{\sqrt{20}}{6}=\frac{\sqrt{5}}{3} \]

Latus rectum:

\[ \frac{2b^2}{a}=\frac{16}{3} \]

Illustration

Exam Significance

• Direct board exam question (formula-based)
• Very common in MCQs (JEE/entrance exams)
• Tests understanding of \(a,b,c\) relationship
• Latus rectum is a frequent trap question

Theory

Standard form of ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(y^2\), major axis is along y-axis
• Foci: \((0,\pm c)\), where \(c=\sqrt{b^2-a^2}\)
• Vertices: \((0,\pm b)\)
• Major axis length: \(2b\)
• Minor axis length: \(2a\)
• Eccentricity: \(e=\frac{c}{b}\)
• Latus rectum: \(\frac{2a^2}{b}\)

Solution Roadmap

• Compare with standard equation
• Identify \(a^2\) and \(b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{b^2-a^2}\)
• Find all required parameters

Solution

Comparing with standard form:

\[ a^2=4,\quad b^2=25 \]

Since the larger denominator is under \(y^2\), the major axis lies along the y-axis.

\[ a=2,\quad b=5 \] \[ c=\sqrt{b^2-a^2}=\sqrt{25-4}=\sqrt{21} \]

Foci:

\[ (0,\pm \sqrt{21}) \]

Vertices:

\[ (0,\pm 5) \]

Length of major axis:

\[ 2b=10 \]

Length of minor axis:

\[ 2a=4 \]

Eccentricity:

\[ e=\frac{\sqrt{21}}{5} \]

Latus rectum:

\[ \frac{2a^2}{b}=\frac{8}{5} \]

Illustration

Exam Significance

• Tests identification of vertical major axis (very common MCQ trap)
• Important for distinguishing between \(a\) and \(b\)
• Latus rectum formula changes with axis orientation
• Frequently appears in board exams and JEE basics

Theory

Standard form of ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(x^2\), major axis lies along x-axis
• Foci: \((\pm c,0)\), where \(c=\sqrt{a^2-b^2}\)
• Vertices: \((\pm a,0)\)
• Major axis length: \(2a\)
• Minor axis length: \(2b\)
• Eccentricity: \(e=\frac{c}{a}\)
• Latus rectum: \(\frac{2b^2}{a}\)

Solution Roadmap

• Compare with standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{a^2-b^2}\)
• Evaluate required parameters

Solution

Comparing with standard form:

\[ a^2=16,\quad b^2=9 \]

Since \(16>9\), the major axis lies along the x-axis.

\[ a=4,\quad b=3 \] \[ c=\sqrt{16-9}=\sqrt{7} \]

Foci:

\[ (\pm \sqrt{7},0) \]

Vertices:

\[ (\pm 4,0) \]

Length of major axis:

\[ 2a=8 \]

Length of minor axis:

\[ 2b=6 \]

Eccentricity:

\[ e=\frac{\sqrt{7}}{4} \]

Latus rectum:

\[ \frac{2b^2}{a}=\frac{9}{2} \]

Illustration

Exam Significance

• Standard horizontal ellipse — very common in board exams
• Tests correct use of \(c=\sqrt{a^2-b^2}\)
• Important for identifying latus rectum orientation
• Forms base for advanced coordinate geometry problems

Theory

Standard form of ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(y^2\), major axis lies along y-axis
• Foci: \((0,\pm c)\), where \(c=\sqrt{b^2-a^2}\)
• Vertices: \((0,\pm b)\)
• Major axis length: \(2b\)
• Minor axis length: \(2a\)
• Eccentricity: \(e=\frac{c}{b}\)
• Latus rectum: \(\frac{2a^2}{b}\)

Solution Roadmap

• Compare with standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{b^2-a^2}\)
• Find all required parameters

Solution

Comparing with standard form:

\[ a^2=25,\quad b^2=100 \]

Since \(100>25\), the major axis lies along the y-axis.

\[ a=5,\quad b=10 \] \[ c=\sqrt{100-25}=\sqrt{75}=5\sqrt{3} \]

Foci:

\[ (0,\pm 5\sqrt{3}) \]

Vertices:

\[ (0,\pm 10) \]

Length of major axis:

\[ 2b=20 \]

Length of minor axis:

\[ 2a=10 \]

Eccentricity:

\[ e=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2} \]

Latus rectum:

\[ \frac{2a^2}{b}=5 \]

Illustration

Exam Significance

• Classic vertical ellipse case (frequent in exams)
• Tests correct identification of \(a\) vs \(b\)
• Important for understanding high eccentricity \(\left(e=\frac{\sqrt{3}}{2}\right)\)
• Useful for graphical interpretation questions

Theory

Standard form of ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(x^2\), major axis lies along x-axis
• Foci: \((\pm c,0)\), where \(c=\sqrt{a^2-b^2}\)
• Vertices: \((\pm a,0)\)
• Major axis length: \(2a\)
• Minor axis length: \(2b\)
• Eccentricity: \(e=\frac{c}{a}\)
• Latus rectum: \(\frac{2b^2}{a}\)

Solution Roadmap

• Compare with standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{a^2-b^2}\)
• Find required parameters

Solution

Comparing with standard form:

\[ a^2=49,\quad b^2=36 \]

Since \(49>36\), the major axis lies along the x-axis.

\[ a=7,\quad b=6 \] \[ c=\sqrt{49-36}=\sqrt{13} \]

Foci:

\[ (\pm \sqrt{13},0) \]

Vertices:

\[ (\pm 7,0) \]

Length of major axis:

\[ 2a=14 \]

Length of minor axis:

\[ 2b=12 \]

Eccentricity:

\[ e=\frac{\sqrt{13}}{7} \]

Latus rectum:

\[ \frac{2b^2}{a}=\frac{72}{7} \]

Illustration

Exam Significance

• Tests strong understanding of horizontal ellipse
• Non-perfect square difference → careful calculation of \(c\)
• Latus rectum often asked in objective exams
• Important for coordinate geometry graph interpretation

Theory

Standard form of ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(y^2\), major axis lies along y-axis
• Foci: \((0,\pm c)\), where \(c=\sqrt{b^2-a^2}\)
• Vertices: \((0,\pm b)\)
• Major axis length: \(2b\)
• Minor axis length: \(2a\)
• Eccentricity: \(e=\frac{c}{b}\)
• Latus rectum: \(\frac{2a^2}{b}\)

Solution Roadmap

• Compare with standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{b^2-a^2}\)
• Find required parameters

Solution

Comparing with standard form:

\[ a^2=100,\quad b^2=400 \]

Since \(400>100\), the major axis lies along the y-axis.

\[ a=10,\quad b=20 \] \[ c=\sqrt{400-100}=\sqrt{300}=10\sqrt{3} \]

Foci:

\[ (0,\pm 10\sqrt{3}) \]

Vertices:

\[ (0,\pm 20) \]

Length of major axis:

\[ 2b=40 \]

Length of minor axis:

\[ 2a=20 \]

Eccentricity:

\[ e=\frac{\sqrt{3}}{2} \]

Latus rectum:

\[ \frac{2a^2}{b}=10 \]

Illustration

Exam Significance

• Classic vertical ellipse with large scale — tests conceptual clarity
• Same eccentricity as Q4 → useful comparison insight
• Helps understand proportional scaling in graphs
• Frequently used in graphical and MCQ-based questions

Theory

To analyze a conic, first convert it into standard form: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(y^2\), major axis lies along y-axis
• Foci: \((0,\pm c)\), where \(c=\sqrt{b^2-a^2}\)
• Vertices: \((0,\pm b)\)
• Major axis length: \(2b\)
• Minor axis length: \(2a\)
• Eccentricity: \(e=\frac{c}{b}\)
• Latus rectum: \(\frac{2a^2}{b}\)

Solution Roadmap

• Convert equation into standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{b^2-a^2}\)
• Find required parameters

Solution

Converting to standard form:

\[ \frac{36x^2}{144}+\frac{4y^2}{144}=1 \] \[ \frac{x^2}{4}+\frac{y^2}{36}=1 \]

Comparing with standard form:

\[ a^2=4,\quad b^2=36 \]

Since \(36>4\), the major axis lies along the y-axis.

\[ a=2,\quad b=6 \] \[ c=\sqrt{36-4}=\sqrt{32}=4\sqrt{2} \]

Foci:

\[ (0,\pm 4\sqrt{2}) \]

Vertices:

\[ (0,\pm 6) \]

Length of major axis:

\[ 2b=12 \]

Length of minor axis:

\[ 2a=4 \]

Eccentricity:

\[ e=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3} \]

Latus rectum:

\[ \frac{2a^2}{b}=\frac{4}{3} \]

Illustration

Exam Significance

• Tests conversion to standard form (very common board question)
• Checks conceptual clarity of vertical ellipse
• Non-perfect square difference → careful simplification needed
• Important for MCQs and coordinate geometry graph questions

Theory

Convert given equation into standard form: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(y^2\), major axis lies along y-axis
• Foci: \((0,\pm c)\), where \(c=\sqrt{b^2-a^2}\)
• Vertices: \((0,\pm b)\)
• Major axis length: \(2b\)
• Minor axis length: \(2a\)
• Eccentricity: \(e=\frac{c}{b}\)
• Latus rectum: \(\frac{2a^2}{b}\)

Solution Roadmap

• Convert equation into standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{b^2-a^2}\)
• Evaluate all required parameters

Solution

Converting to standard form:

\[ 16x^2 + y^2 = 16 \] \[ \frac{16x^2}{16} + \frac{y^2}{16} = 1 \] \[ x^2 + \frac{y^2}{16} = 1 \]

Comparing with standard form:

\[ a^2=1,\quad b^2=16 \]

Since \(16>1\), the major axis lies along the y-axis.

\[ a=1,\quad b=4 \] \[ c=\sqrt{16-1}=\sqrt{15} \]

Foci:

\[ (0,\pm \sqrt{15}) \]

Vertices:

\[ (0,\pm 4) \]

Length of major axis:

\[ 2b=8 \]

Length of minor axis:

\[ 2a=2 \]

Eccentricity:

\[ e=\frac{\sqrt{15}}{4} \]

Latus rectum:

\[ \frac{2a^2}{b}=\frac{1}{2} \]

Illustration

Exam Significance

• Very important for testing normalization step
• Highly eccentric ellipse → visually narrow shape
• Small latus rectum → common conceptual confusion
• Frequently appears in MCQs and graphical interpretation

Theory

Convert the equation into standard form: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• If larger denominator is under \(x^2\), major axis lies along x-axis
• Foci: \((\pm c,0)\), where \(c=\sqrt{a^2-b^2}\)
• Vertices: \((\pm a,0)\)
• Major axis length: \(2a\)
• Minor axis length: \(2b\)
• Eccentricity: \(e=\frac{c}{a}\)
• Latus rectum: \(\frac{2b^2}{a}\)

Solution Roadmap

• Convert into standard form
• Identify \(a^2, b^2\)
• Determine axis orientation
• Compute \(c=\sqrt{a^2-b^2}\)
• Find required parameters

Solution

Converting to standard form:

\[ \frac{4x^2}{36}+\frac{9y^2}{36}=1 \] \[ \frac{x^2}{9}+\frac{y^2}{4}=1 \]

Comparing with standard form:

\[ a^2=9,\quad b^2=4 \]

Since \(9>4\), the major axis lies along the x-axis.

\[ a=3,\quad b=2 \] \[ c=\sqrt{9-4}=\sqrt{5} \]

Foci:

\[ (\pm \sqrt{5},0) \]

Vertices:

\[ (\pm 3,0) \]

Length of major axis:

\[ 2a=6 \]

Length of minor axis:

\[ 2b=4 \]

Eccentricity:

\[ e=\frac{\sqrt{5}}{3} \]

Latus rectum:

\[ \frac{2b^2}{a}=\frac{8}{3} \]

Illustration

Exam Significance

• Tests conversion to standard form (very common)
• Balanced ellipse (a and b close) → moderate eccentricity
• Good comparison with Q8 (contrast in shape)
• Frequently used in coordinate geometry MCQs

Theory

For an ellipse centered at origin with major axis along x-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

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

Solution Roadmap

• Identify \(a\) from vertices
• Identify \(c\) from foci
• Use relation \(c^2 = a^2 - b^2\)
• Find \(b^2\)
• Substitute into standard equation

Solution

Given:

\[ a=5,\quad c=4 \]

Since vertices and foci lie on x-axis, the major axis is along x-axis.

\[ c^2 = a^2 - b^2 \] \[ 16 = 25 - b^2 \] \[ b^2 = 9 \]

Equation of ellipse:

\[ \frac{x^2}{25}+\frac{y^2}{9}=1 \]

Illustration

Exam Significance

• Very common “reverse problem” in board exams
• Tests correct use of relation \(c^2 = a^2 - b^2\)
• Students often confuse \(a\) and \(c\) → high scoring trap
• Foundation for advanced conic construction problems

Theory

For an ellipse centered at origin with major axis along y-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

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

Solution Roadmap

• Identify \(b\) from vertices
• Identify \(c\) from foci
• Use relation \(c^2 = b^2 - a^2\)
• Find \(a^2\)
• Substitute into standard equation

Solution

Given:

\[ b=13,\quad c=5 \]

Since vertices and foci lie on y-axis, the major axis is along the y-axis.

\[ c^2 = b^2 - a^2 \] \[ 25 = 169 - a^2 \] \[ a^2 = 144 \]

Equation of ellipse:

\[ \frac{x^2}{144}+\frac{y^2}{169}=1 \]

Illustration

Exam Significance

• Tests reverse construction of ellipse (very common in boards)
• Checks correct use of \(c^2 = b^2 - a^2\)
• Important to distinguish vertical vs horizontal ellipse
• Foundation for coordinate geometry and conic modelling

Theory

For an ellipse centered at origin with major axis along x-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

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

Solution Roadmap

• Identify \(a\) from vertices
• Identify \(c\) from foci
• Use relation \(c^2 = a^2 - b^2\)
• Find \(b^2\)
• Substitute into standard equation

Solution

Given:

\[ a=6,\quad c=4 \]

Since vertices and foci lie on x-axis, the major axis is along x-axis.

\[ c^2 = a^2 - b^2 \] \[ 16 = 36 - b^2 \] \[ b^2 = 20 \]

Equation of ellipse:

\[ \frac{x^2}{36}+\frac{y^2}{20}=1 \]

Illustration

Exam Significance

• Classic reverse problem — very common in exams
• Tests correct application of \(c^2 = a^2 - b^2\)
• Non-perfect square \(b^2\) → requires careful handling
• Important for developing geometric intuition

Theory

For an ellipse centered at origin: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• Ends of major axis: \((\pm a,0)\)
• Ends of minor axis: \((0,\pm b)\)
• If major axis is along x-axis → use standard form directly

Solution Roadmap

• Identify \(a\) from major axis endpoints
• Identify \(b\) from minor axis endpoints
• Confirm axis orientation
• Substitute into standard equation

Solution

Given:

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

Since the major axis lies along the x-axis, the ellipse is centered at origin.

Equation of ellipse:

\[ \frac{x^2}{9}+\frac{y^2}{4}=1 \]

Illustration

Exam Significance

• Direct concept question — very common in board exams
• No need to compute \(c\) → faster solving
• Tests identification of axes from endpoints
• Important for quick MCQ solving

Theory

For an ellipse centered at origin with major axis along y-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• Ends of major axis: \((0,\pm b)\)
• Ends of minor axis: \((\pm a,0)\)
• If major axis is vertical → denominator under \(y^2\) is larger

Solution Roadmap

• Identify \(b\) from major axis endpoints
• Identify \(a\) from minor axis endpoints
• Confirm axis orientation
• Substitute into standard equation

Solution

From given data:

\[ b=\sqrt{5},\quad a=1 \] \[ b^2=5,\quad a^2=1 \]

Since the major axis lies along the y-axis, use vertical ellipse form.

Equation of ellipse:

\[ \frac{x^2}{1}+\frac{y^2}{5}=1 \]

Illustration

Exam Significance

• Tests handling of irrational values like \(\sqrt{5}\)
• Important for identifying vertical ellipse correctly
• Direct substitution problem → quick scoring
• Common in board exams and MCQs

Theory

For an ellipse centered at origin with major axis along x-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

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

Solution Roadmap

• Find \(a\) from length of major axis
• Identify \(c\) from foci
• Use relation \(c^2 = a^2 - b^2\)
• Find \(b^2\)
• Substitute into equation

Solution

Given:

\[ 2a=26 \Rightarrow a=13,\quad c=5 \]

Since foci lie on x-axis, the major axis is along x-axis.

\[ c^2 = a^2 - b^2 \] \[ 25 = 169 - b^2 \] \[ b^2 = 144 \]

Equation of ellipse:

\[ \frac{x^2}{169}+\frac{y^2}{144}=1 \]

Illustration

Exam Significance

• Combines axis length + foci → very common exam pattern
• Tests understanding of \(2a\) vs \(a\)
• Recognizable Pythagorean triple (5,12,13) → speeds up solving
• Important for both board exams and MCQs

Theory

For an ellipse centered at origin with major axis along y-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

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

Solution Roadmap

• Find \(a\) from minor axis length
• Identify \(c\) from foci
• Determine axis orientation
• Use \(c^2 = b^2 - a^2\)
• Find \(b^2\) and form equation

Solution

Given:

\[ 2a=16 \Rightarrow a=8,\quad c=6 \]

Since foci lie on y-axis, the major axis is along the y-axis.

\[ c^2 = b^2 - a^2 \] \[ 36 = b^2 - 64 \] \[ b^2 = 100 \]

Equation of ellipse:

\[ \frac{x^2}{64}+\frac{y^2}{100}=1 \]

Illustration

Exam Significance

• Tests understanding of minor axis vs major axis distinction
• Common mistake: students confuse \(2a\) with \(2b\)
• Vertical ellipse identification is crucial
• Forms base for higher-level conic problems

Theory

For an ellipse centered at origin with major axis along x-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• Foci: \((\pm c,0)\)
• \(a\) is semi-major axis
• Relation: \(c^2 = a^2 - b^2\)

Solution Roadmap

• Identify \(c\) from foci
• Use given \(a\)
• Apply \(c^2 = a^2 - b^2\)
• Find \(b^2\)
• Substitute into standard equation

Solution

Given:

\[ a=4,\quad c=3 \]

Since foci lie on x-axis, the major axis is along x-axis.

\[ c^2 = a^2 - b^2 \] \[ 9 = 16 - b^2 \] \[ b^2 = 7 \]

Equation of ellipse:

\[ \frac{x^2}{16}+\frac{y^2}{7}=1 \]

Illustration

Exam Significance

• Tests direct use of \(c^2 = a^2 - b^2\)
• Common MCQ pattern: given \(a\) and foci
• Non-integer \(b\) → requires precision
• Important for building equation quickly

Theory

For an ellipse centered at origin with major axis along x-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• Foci: \((\pm c,0)\)
• \(a\) is semi-major axis
• Relation: \(c^2 = a^2 - b^2\)

Solution Roadmap

• Identify axis orientation from foci
• Use given \(b\) and \(c\)
• Apply \(c^2 = a^2 - b^2\)
• Find \(a^2\)
• Form equation

Solution

Given:

\[ b=3,\quad c=4 \]

Since foci lie on x-axis, the major axis is along x-axis.

\[ c^2 = a^2 - b^2 \] \[ 16 = a^2 - 9 \] \[ a^2 = 25 \]

Equation of ellipse:

\[ \frac{x^2}{25}+\frac{y^2}{9}=1 \]

Illustration

Exam Significance

• Given \(b\) and \(c\) → slightly uncommon but important pattern
• Tests correct rearrangement of \(c^2 = a^2 - b^2\)
• Recognizable 3–4–5 triangle → speeds solving
• Frequently appears in MCQs

Theory

For an ellipse centered at origin with major axis along y-axis: \[ \frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \]

• Unknowns: \(a^2, b^2\)
• Points on ellipse satisfy the equation
• Use simultaneous equations to solve

Solution Roadmap

• Write standard equation (vertical ellipse)
• Substitute given points
• Form two equations
• Solve using substitution/elimination
• Construct final equation

Solution

Standard form:

\[ \frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \]

Substituting point (3,2):

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

Substituting point (1,6):

\[ \frac{1}{b^2}+\frac{36}{a^2}=1 \]

Let:

\[ x=\frac{1}{b^2},\quad y=\frac{1}{a^2} \] \[ 9x+4y=1 \] \[ x+36y=1 \]

Multiply second equation by 9:

\[ 9x+324y=9 \]

Subtract:

\[ -320y=-8 \Rightarrow y=\frac{1}{40} \] \[ x+36\left(\frac{1}{40}\right)=1 \] \[ x=\frac{1}{10} \] \[ a^2=40,\quad b^2=10 \]

Equation of ellipse:

\[ \frac{x^2}{10}+\frac{y^2}{40}=1 \]

Illustration

Exam Significance

• Most important type: ellipse through given points
• Tests ability to form and solve simultaneous equations
• Substitution trick \(x=\frac{1}{b^2}, y=\frac{1}{a^2}\) saves time
• Very common in board exams and competitive exams

Theory

For an ellipse centered at origin with major axis along x-axis: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

• Unknowns: \(a^2, b^2\)
• Points on ellipse satisfy the equation
• Use simultaneous equations to determine parameters

Solution Roadmap

• Write standard equation (horizontal ellipse)
• Substitute given points
• Form two equations
• Solve using substitution/elimination
• Construct final equation

Solution

Standard form:

\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

Substituting (4,3):

\[ \frac{16}{a^2}+\frac{9}{b^2}=1 \]

Substituting (6,2):

\[ \frac{36}{a^2}+\frac{4}{b^2}=1 \]

Let:

\[ x=\frac{1}{a^2},\quad y=\frac{1}{b^2} \] \[ 16x+9y=1 \] \[ 36x+4y=1 \]

Multiply:

\[ 64x+36y=4 \] \[ 324x+36y=9 \]

Subtract:

\[ -260x=-5 \Rightarrow x=\frac{1}{52} \] \[ 16\left(\frac{1}{52}\right)+9y=1 \] \[ y=\frac{1}{13} \] \[ a^2=52,\quad b^2=13 \]

Equation of ellipse:

\[ \frac{x^2}{52}+\frac{y^2}{13}=1 \]

Illustration

Exam Significance

• Most advanced type in this exercise
• Tests full command over ellipse equation formation
• Requires solving simultaneous equations efficiently
• Very important for board exams and competitive exams