Quick answer: The ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 a > b a > b a > b e = 1 − b 2 a 2 e = \sqrt{1 - \frac{b^2}{a^2}} e = 1 − a 2 b 2 0 < e < 1 0 < e < 1 0 < e < 1 ( ± a e , 0 ) (\pm ae, 0) ( ± a e , 0 ) x = ± a e x = \pm\frac{a}{e} x = ± e a 2 b 2 a \frac{2b^2}{a} a 2 b 2 2 a 2a 2 a y = m x + c y = mx + c y = m x + c c 2 = a 2 m 2 + b 2 c^2 = a^2m^2 + b^2 c 2 = a 2 m 2 + b 2
The parabola has e = 1 e = 1 e = 1 parabola standard forms ); an ellipse has e < 1 e < 1 e < 1
Standard ellipse (a > b)
Item Value Equation x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 Centre ( 0 , 0 ) (0, 0) ( 0 , 0 ) Vertices ( ± a , 0 ) (\pm a, 0) ( ± a , 0 ) Major / minor axis 2 a 2a 2 a 2 b 2b 2 b Eccentricity e = 1 − b 2 a 2 e = \sqrt{1 - \frac{b^2}{a^2}} e = 1 − a 2 b 2 b 2 = a 2 ( 1 − e 2 ) b^2 = a^2(1 - e^2) b 2 = a 2 ( 1 − e 2 ) Foci ( ± a e , 0 ) (\pm ae, 0) ( ± a e , 0 ) Directrices x = ± a e x = \pm\frac{a}{e} x = ± e a Latus rectum 2 b 2 a \frac{2b^2}{a} a 2 b 2 Parametric point ( a cos θ , b sin θ ) (a\cos\theta, b\sin\theta) ( a cos θ , b sin θ )
Focal distances: for P ( x 1 , y 1 ) P(x_1, y_1) P ( x 1 , y 1 ) S P = a − e x 1 SP = a - ex_1 S P = a − e x 1 S ′ P = a + e x 1 S'P = a + ex_1 S ′ P = a + e x 1 S P + S ′ P = 2 a SP + S'P = 2a S P + S ′ P = 2 a
When b > a
The major axis is along the y-axis: e = 1 − a 2 b 2 e = \sqrt{1 - \frac{a^2}{b^2}} e = 1 − b 2 a 2 ( 0 , ± b e ) (0, \pm be) ( 0 , ± b e ) y = ± b e y = \pm\frac{b}{e} y = ± e b 2 a 2 b \frac{2a^2}{b} b 2 a 2
Tangents and normals
Form Equation Tangent at ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) x x 1 a 2 + y y 1 b 2 = 1 \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 a 2 x x 1 + b 2 y y 1 = 1 Tangent at θ x cos θ a + y sin θ b = 1 \frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1 a x c o s θ + b y s i n θ = 1 Slope form y = m x ± a 2 m 2 + b 2 y = mx \pm\sqrt{a^2m^2 + b^2} y = m x ± a 2 m 2 + b 2 Normal at ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) a 2 x x 1 − b 2 y y 1 = a 2 − b 2 \frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 x 1 a 2 x − y 1 b 2 y = a 2 − b 2
Condition of tangency: c 2 = a 2 m 2 + b 2 c^2 = a^2m^2 + b^2 c 2 = a 2 m 2 + b 2
Director circle: perpendicular tangents meet on x 2 + y 2 = a 2 + b 2 x^2 + y^2 = a^2 + b^2 x 2 + y 2 = a 2 + b 2
Auxiliary circle: x 2 + y 2 = a 2 x^2 + y^2 = a^2 x 2 + y 2 = a 2
Position of a point
Let S 1 = x 1 2 a 2 + y 1 2 b 2 − 1 S_1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1 S 1 = a 2 x 1 2 + b 2 y 1 2 − 1 S 1 < 0 S_1 < 0 S 1 < 0 S 1 = 0 S_1 = 0 S 1 = 0 S 1 > 0 S_1 > 0 S 1 > 0
Worked examples
Example 1: all the basics
For x 2 25 + y 2 16 = 1 \frac{x^2}{25} + \frac{y^2}{16} = 1 25 x 2 + 16 y 2 = 1
Step 1: a = 5 a = 5 a = 5 b = 4 b = 4 b = 4 e = 1 − 16 25 = 3 5 e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5} e = 1 − 25 16 = 5 3
Step 2: Foci ( ± 3 , 0 ) (\pm 3, 0) ( ± 3 , 0 ) x = ± 25 3 x = \pm\frac{25}{3} x = ± 3 25 2 ⋅ 16 5 = 32 5 \frac{2 \cdot 16}{5} = \frac{32}{5} 5 2 ⋅ 16 = 5 32
Example 2: equation from data
Find the ellipse with foci ( ± 4 , 0 ) (\pm 4, 0) ( ± 4 , 0 ) 4 5 \frac{4}{5} 5 4
Step 1: a e = 4 ae = 4 a e = 4 a = 5 a = 5 a = 5
Step 2: b 2 = a 2 − a 2 e 2 = 25 − 16 = 9 b^2 = a^2 - a^2e^2 = 25 - 16 = 9 b 2 = a 2 − a 2 e 2 = 25 − 16 = 9
Answer: x 2 25 + y 2 9 = 1 \frac{x^2}{25} + \frac{y^2}{9} = 1 25 x 2 + 9 y 2 = 1
Example 3: completing the square
Find the centre and eccentricity of 4 x 2 + 9 y 2 − 16 x + 18 y − 11 = 0 4x^2 + 9y^2 - 16x + 18y - 11 = 0 4 x 2 + 9 y 2 − 16 x + 18 y − 11 = 0
Step 1: 4 ( x − 2 ) 2 + 9 ( y + 1 ) 2 = 11 + 16 + 9 = 36 4(x - 2)^2 + 9(y + 1)^2 = 11 + 16 + 9 = 36 4 ( x − 2 ) 2 + 9 ( y + 1 ) 2 = 11 + 16 + 9 = 36
Step 2: ( x − 2 ) 2 9 + ( y + 1 ) 2 4 = 1 \frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1 9 ( x − 2 ) 2 + 4 ( y + 1 ) 2 = 1 ( 2 , − 1 ) (2, -1) ( 2 , − 1 ) e = 1 − 4 9 = 5 3 e = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3} e = 1 − 9 4 = 3 5
Example 4: tangent condition
Find the tangents of slope 1 to x 2 16 + y 2 9 = 1 \frac{x^2}{16} + \frac{y^2}{9} = 1 16 x 2 + 9 y 2 = 1
c 2 = 16 + 9 = 25 c^2 = 16 + 9 = 25 c 2 = 16 + 9 = 25 y = x ± 5 y = x \pm 5 y = x ± 5
Summary table
Item Result (a > b a > b a > b b 2 b^2 b 2 a 2 ( 1 − e 2 ) a^2(1 - e^2) a 2 ( 1 − e 2 ) Foci ( ± a e , 0 ) (\pm ae, 0) ( ± a e , 0 ) Directrices x = ± a e x = \pm\frac{a}{e} x = ± e a Latus rectum 2 b 2 a \frac{2b^2}{a} a 2 b 2 S P + S ′ P SP + S'P S P + S ′ P 2 a 2a 2 a Tangency c 2 = a 2 m 2 + b 2 c^2 = a^2m^2 + b^2 c 2 = a 2 m 2 + b 2 Director circle x 2 + y 2 = a 2 + b 2 x^2 + y^2 = a^2 + b^2 x 2 + y 2 = a 2 + b 2
Exam tips
Check which denominator is larger before writing the foci.
Use S P + S ′ P = 2 a SP + S'P = 2a S P + S ′ P = 2 a
Complete the square to bring a general ellipse to standard form.
The distance between the foci is 2 a e 2ae 2 a e 2 a e \frac{2a}{e} e 2 a
Next, read hyperbola: eccentricity, asymptotes and tangents , or see the NIMCET Coordinate Geometry and Vectors chapter .
Common mistakes
Using e = 1 + b 2 a 2 e = \sqrt{1 + \frac{b^2}{a^2}} e = 1 + a 2 b 2
Putting the foci on the x-axis when b > a b > a b > a
Writing the latus rectum as 2 a 2 b \frac{2a^2}{b} b 2 a 2 a > b a > b a > b
Forgetting the ± in the slope form of the tangent.
Frequently Asked Questions
What is the eccentricity of an ellipse?
For x²/a² + y²/b² = 1 with a > b, e = √(1 − b²/a²), and it always lies between 0 and 1.
What are the foci of an ellipse?
For a > b, the foci are (±ae, 0).
What is the length of the latus rectum of an ellipse?
It is 2b²/a when a > b.
What is the sum of focal distances of a point on an ellipse?
It equals 2a, the length of the major axis.
What is the condition for y = mx + c to touch an ellipse?
c² = a²m² + b².
