Parabola: standard forms
The reflector of a torch and a satellite dish have the shape of a parabola. All the waves meet at one special point of the dish; that point is the focus. Building on classes 9–10 : In classes 9–10 we learnt the distance between two points and the circle; a parabola is also made from a condition on distance: "equal distance from the focus and the directrix".
The locus of points at equal distance from a fixed point (the focus) and a fixed line (the directrix) is a parabola. The line through the focus perpendicular to the directrix is the axis; the point where the axis meets the parabola is the vertex.
Rule 1: The four standard forms — y² = 4ax: focus (a, 0), directrix x = −a. y² = −4ax: focus (−a, 0), directrix x = a. x² = 4ay: focus (0, a), directrix y = −a. x² = −4ay: focus (0, −a), directrix y = a. The vertex is the origin.
In the figure : The distances from P to the focus F and to the point M on the directrix are equal (PF = PM); the green segment is the latus rectum.
Rule 2: Latus rectum and focal distance — The chord through the focus perpendicular to the axis is the latus rectum; for y² = 4ax its length is 4a and its ends are (a, ±2a). The distance of a point (x₁, y₁) from the focus is x₁ + a.
How is the formula obtained? : Let P(x, y) lie on the parabola, with focus F(a, 0) and directrix x = −a. PF = PM means √((x − a)² + y²) = x + a. Squaring both sides: (x − a)² + y² = (x + a)². The difference of (x + a)² and (x − a)² is 4ax, so y² = 4ax.
- Easy : y² = 12x: 4a = 12, a = 3. The focus is (3, 0), the directrix x = −3 and the latus rectum has length 12.
- Medium : Focus (0, −5) and vertex at the origin: a = 5 and the parabola opens downward. The equation is x² = −4ay = −20y.
- Exam-type : A point on y² = 16x is at distance 9 from the focus: x₁ + 4 = 9, x₁ = 5. y² = 80, y = ±4√5. The points are (5, 4√5) and (5, −4√5).
If the form has y², the parabola opens left or right (axis X); if it has x², it opens up or down (axis Y). A positive coefficient means right or up, negative means left or down.
Taking a = 12 from y² = 12x. Since 4a = 12, a = 3. The focus is (3, 0), not (12, 0).
Divide the coefficient by 4: that is a. For y² = 4ax the focus = (coefficient ÷ 4, 0), the directrix is x = −(coefficient ÷ 4), and the latus rectum = the coefficient itself. Memory clue : "divide by four to get a; the latus rectum is the coefficient".
CET gives the ends of the latus rectum; take a as their x-coordinate and 4a is the coefficient.
CET-style example : The ends of the latus rectum of a parabola (vertex at the origin) are (3, 6) and (3, −6). Its equation is? (A) y² = 12x (B) y² = 6x (C) y² = 24x (D) y² = 3x
Answer : (A). The latus rectum is the line x = a, so a = 3 and y² = 4 × 3 × x = 12x. Time-saving method : End point (a, 2a): 2a = 6 and a = 3; 4a = 12.
Spot the difference
| Point | y² = 4ax | x² = 4ay |
|---|---|---|
| Axis | X-axis | Y-axis |
| Focus | (a, 0) | (0, a) |
| Directrix | x = −a | y = −a |
| Opens | to the right | upward |
Practice questions
- 1Write the focus and directrix of y² = 20x.
- 2State the length of the latus rectum of x² = −16y.
- 3Write the equation of the parabola with focus (6, 0) and vertex at the origin.
- 4The parabola x² = 4ay passes through (4, 2). Write its equation.
- 5Find the distance of the point (3, 6) on y² = 12x from the focus.
- 6Write the equation of the parabola with focus (−4, 0) and vertex at the origin.
- 7Points on y² = 8x are at distance 10 from the focus. Find them.
- 8Find the area of the triangle formed by the vertex and the ends of the latus rectum of y² = 8x.
Show answers (try them yourself first)
- 1a = 5. The focus is (5, 0) and the directrix is x = −5.
- 2The length is 16.
- 3a = 6, so y² = 24x.
- 416 = 4a × 2 = 8a, so a = 2 and x² = 8y.
- 5a = 3. The distance = x₁ + a = 3 + 3 = 6.
- 6a = 4 and the parabola opens to the left: y² = −16x.
- 7a = 2. x₁ + 2 = 10, x₁ = 8, y² = 64, y = ±8. The points are (8, 8) and (8, −8).
- 8a = 2. The ends are (2, 4) and (2, −4); base 8 and height 2. Area = ½ × 8 × 2 = 8.
Parabola: parametric form and tangents
When a ball is thrown it is at a fixed place at a fixed instant; as the instant changes the place changes. Points that depend on one variable (the instant) in this way are the parametric form. Building on classes 9–10 : In classes 9–10 we learnt the slope of a line and its equation; a tangent is a line that meets the parabola at exactly one point.
Every point of y² = 4ax can be written as (at², 2at); t is the parameter. A line that touches the parabola at exactly one point is a tangent.
Rule 1: Tangent at a point — To y² = 4ax at (x₁, y₁): yy₁ = 2a(x + x₁). At the point t: ty = x + at². To x² = 4ay at (x₁, y₁): xx₁ = 2a(y + y₁).
In the figure : The tangent at P(at², 2at) meets the parabola only at P.
Rule 2: Tangent from the slope — y = mx + c touches y² = 4ax if c = a/m. The point of contact is (a/m², 2a/m). For x² = 4ay the condition is c = −am².
How is the formula obtained? : Put y = mx + c in y² = 4ax: m²x² + (2mc − 4a)x + c² = 0. For a tangent there must be only one root, so the discriminant is zero: (2mc − 4a)² − 4m²c² = 0. Expanding: 16a² − 16amc = 0, so c = a/m. The double root is x = −(2mc − 4a)/(2m²) = a/m², so the point of contact is (a/m², 2a/m).
- Easy : On y² = 4x (a = 1) with t = 3: (at², 2at) = (1 × 9, 2 × 1 × 3) = (9, 6).
- Medium : Tangent to y² = 16x at (4, 8): a = 4, 8y = 8(x + 4), so x − y + 4 = 0.
- Exam-type : y = 3x + c touches y² = 24x: a = 6 and m = 3, so c = a/m = 2.
yy₁ = 2a(x + x₁) is a tangent only if (x₁, y₁) lies on the parabola. For an outside point the same equation gives the chord of contact.
Writing c = am instead of c = a/m. a is divided by m, not multiplied.
To write a tangent replace y² by yy₁ and 2x by x + x₁ (so 4ax becomes 2a(x + x₁)). For x² = 4ay replace x² by xx₁ and 2y by y + y₁. Memory clue : "split the square: one full, one half".
When CET says "the line touches", bring the line to y = mx + c and use c = a/m; avoid the long discriminant work.
CET-style example : The line y = x + k touches y² = 36x. k = ? (A) 36 (B) 18 (C) 9 (D) 4
Answer : (C). 4a = 36, a = 9 and m = 1; k = a/m = 9. Time-saving method : Coefficient ÷ 4 = a; then a ÷ m.
Spot the difference
| Point | Tangent at a point | Tangent by slope |
|---|---|---|
| Given | point (x₁, y₁) | slope m |
| Formula | yy₁ = 2a(x + x₁) | y = mx + a/m |
| How many | one | one (for y² = 4ax) |
Practice questions
- 1Write the point with t = −2 on y² = 4x.
- 2Write the point with t = 2 on y² = 16x.
- 3y = 2x + c touches y² = 16x. Find c.
- 4Write the tangent to y² = 12x at (3, −6).
- 5Write the tangent to x² = 12y at (6, 3).
- 6Write the tangent of slope 1/2 to y² = 8x.
- 7At which point does y = x + 4 touch y² = 16x?
- 8Write the tangent to y² = 8x at the point t = 3.
Show answers (try them yourself first)
- 1(4, −4).
- 2a = 4. The point is (16, 16).
- 3a = 4, c = a/m = 2.
- 4a = 3. −6y = 6(x + 3), so x + y + 3 = 0.
- 5a = 3. 6x = 6(y + 3), so x − y − 3 = 0.
- 6a = 2, c = a/m = 4. y = x/2 + 4, i.e. x − 2y + 8 = 0.
- 7a = 4, m = 1. The point of contact (a/m², 2a/m) = (4, 8).
- 8a = 2. ty = x + at²: 3y = x + 18, i.e. x − 3y + 18 = 0.
Ellipse: standard form
A round plate looks like a long oval when viewed from a tilt; that is the shape of an ellipse. The orbits of planets are also elliptical. Building on classes 9–10 : In classes 9–10 we learnt the circle (equal distance from one centre); an ellipse has two foci and the sum of the distances stays constant.
The locus of points for which the sum of the distances from two fixed points (the foci) is a constant 2a is an ellipse. The constant 2a must be greater than the distance between the foci. The major axis is 2a and the minor axis is 2b.
Rule 1: Standard equation — x²/a² + y²/b² = 1 (a > b). b² = a²(1 − e²), i.e. c² = a² − b² and e = c/a. The foci are (±ae, 0), the directrices x = ±a/e, e < 1. If the major axis is along the Y-axis the equation is x²/b² + y²/a² = 1 and the foci are (0, ±ae).
In the figure : The distances from P to the two foci add up to 2a.
Rule 2: Latus rectum and sum of focal distances — The length of the latus rectum is 2b²/a. For any point on the ellipse PF₁ + PF₂ = 2a. The distance between the foci is 2ae.
How is the formula obtained? : The foci are (±c, 0) and PF₁ + PF₂ = 2a. So √((x + c)² + y²) + √((x − c)² + y²) = 2a. Moving one root to the other side, squaring and simplifying gives (a² − c²)x² + a²y² = a²(a² − c²). Putting b² = a² − c² gives b²x² + a²y² = a²b²; dividing both sides by a²b² gives x²/a² + y²/b² = 1.
- Easy : x²/36 + y²/20 = 1: a = 6, b² = 20, c² = 36 − 20 = 16, c = 4. The foci are (±4, 0) and e = 4/6 = 2/3.
- Medium : Foci (±2, 0) and e = 1/2: ae = 2 so a = 4. b² = a² − c² = 16 − 4 = 12. The equation is x²/16 + y²/12 = 1.
- Exam-type : 4x² + 9y² = 144: divide by 144, x²/36 + y²/16 = 1. a = 6, b = 4, c² = 20, e = √20/6 = √5/3. Latus rectum = 2b²/a = 32/6 = 16/3.
In an ellipse the larger denominator is a², and the variable it sits under decides the major axis. Larger denominator under x² means the major axis is along the X-axis.
Writing c² = a² + b² (that is the hyperbola formula). In an ellipse c² = a² − b².
Larger denominator − smaller denominator = c². E.g. 25 and 9: c² = 16, c = 4; e = c ÷ a = 4/5. Memory clue : "ellipse subtracts, hyperbola adds".
CET gives the equation with coefficients, like 4x² + 9y² = 144; first divide until the right side is 1 and only then read a² and b².
CET-style example : For the ellipse 9x² + 16y² = 144 the sum of the distances of any point from the two foci is? (A) 6 (B) 8 (C) 12 (D) 16
Answer : (B). Divide by 144: x²/16 + y²/9 = 1, a = 4. The sum = 2a = 8. Time-saving method : Square root of the larger denominator 16 is 4, then double it: 8.
Spot the difference
| Point | Circle | Ellipse |
|---|---|---|
| Foci | one centre | two foci |
| Equation | x² + y² = r² | x²/a² + y²/b² = 1 |
| Eccentricity | 0 | between 0 and 1 |
Practice questions
- 1Find a, b and e for x²/100 + y²/64 = 1.
- 2Find the foci of x²/169 + y²/25 = 1.
- 3Find the lengths of the major and minor axes of x²/49 + y²/25 = 1.
- 4Find the eccentricity and the length of the latus rectum of 16x² + 25y² = 400.
- 5Write the equation of the ellipse with foci (±5, 0) and minor axis of length 24.
- 6An ellipse with its centre at the origin (axes along the coordinate axes) passes through (4, 0) and (0, 3). Write its equation.
- 7Write the equation of the ellipse with eccentricity 4/5 and foci (±8, 0).
- 8Find the distance between the two directrices of x²/16 + y²/7 = 1.
Show answers (try them yourself first)
- 1a = 10, b = 8, c² = 36 so c = 6 and e = 3/5.
- 2c² = 169 − 25 = 144, c = 12. The foci are (±12, 0).
- 3The major axis is 14 and the minor axis is 10.
- 4x²/25 + y²/16 = 1; c² = 9, e = 3/5. Latus rectum = 2 × 16/5 = 32/5.
- 5b = 12, c = 5, a² = 144 + 25 = 169. The equation is x²/169 + y²/144 = 1.
- 6a = 4 and b = 3: x²/16 + y²/9 = 1.
- 7ae = 8 and e = 4/5 so a = 10. b² = 100 − 64 = 36. The equation is x²/100 + y²/36 = 1.
- 8c² = 16 − 7 = 9, c = 3, e = 3/4. The distance = 2a/e = 8 ÷ 3/4 = 32/3.
Ellipse: parametric form, tangents and special circles
Every point on a round disc can be fixed by an angle; if the disc is squeezed from one side the same point lands on an ellipse. So the points of an ellipse are also fixed by one angle (θ). Building on classes 9–10 : In classes 9–10 we learnt cos θ and sin θ and cos²θ + sin²θ = 1; the parametric form of the ellipse comes from it.
A point on the ellipse can be written as (a cos θ, b sin θ); θ is the parameter. The circle x² + y² = a² of radius a is the auxiliary circle. The locus of points from which two tangents are perpendicular is the director circle.
Rule 1: Tangents — At (x₁, y₁): xx₁/a² + yy₁/b² = 1. At θ: x cos θ/a + y sin θ/b = 1. For slope m: y = mx ± √(a²m² + b²); the condition of tangency is c² = a²m² + b².
In the figure : Q on the auxiliary circle and P on the ellipse have the same x-coordinate (a cos θ).
Rule 2: Auxiliary and director circles — Auxiliary circle: x² + y² = a². Director circle: x² + y² = a² + b². The two tangents from a point on the director circle are perpendicular.
How is the formula obtained? : Put y = mx + c in x²/a² + y²/b² = 1: (b² + a²m²)x² + 2a²mcx + a²(c² − b²) = 0. The discriminant is zero: 4a⁴m²c² − 4(b² + a²m²)a²(c² − b²) = 0. Dividing by a² and expanding gives −b²c² + b⁴ + a²m²b² = 0, so c² = a²m² + b².
- Easy : On x²/16 + y²/9 = 1 with θ = 30°: (4 cos 30°, 3 sin 30°) = (4 × √3/2, 3 × 1/2) = (2√3, 3/2).
- Medium : Tangent to x²/20 + y²/5 = 1 at (4, 1): 4x/20 + 1·y/5 = 1, i.e. x/5 + y/5 = 1, so x + y − 5 = 0.
- Exam-type : y = 2x + c touches x²/9 + y²/64 = 1: c² = a²m² + b² = 9 × 4 + 64 = 100, so c = ±10 and the lines are 2x − y ± 10 = 0.
For a given slope an ellipse always has two parallel tangents (±), one on each side.
Writing c² = a²m² − b² in the condition of tangency (that is for the hyperbola). The ellipse has +.
The radius squared of the director circle = the sum of both denominators. For x²/25 + y²/144: 25 + 144 = 169. Memory clue : "director = add the denominators".
When CET asks for "the locus of the intersection of perpendicular tangents" the answer is the director circle; the auxiliary circle (x² + y² = a²) is the wrong option.
CET-style example : The locus of the intersection of perpendicular tangents to x²/25 + y²/144 = 1 is? (A) x² + y² = 144 (B) x² + y² = 169 (C) x² + y² = 25 (D) x² + y² = 119
Answer : (B). The director circle x² + y² = a² + b² = 25 + 144 = 169. Time-saving method : Add the two denominators: 25 + 144.
Spot the difference
| Point | Auxiliary circle | Director circle |
|---|---|---|
| Equation | x² + y² = a² | x² + y² = a² + b² |
| Radius | a | √(a² + b²) |
| Use | showing the parameter θ | locus of perpendicular tangents |
Practice questions
- 1Write the point with θ = 90° on x²/36 + y²/25 = 1.
- 2Write the auxiliary circle of x²/81 + y²/49 = 1.
- 3Write the director circle of x²/9 + y²/16 = 1.
- 4Find the point with θ = 60° on x²/16 + y²/36 = 1.
- 5Write the tangent to x²/20 + y²/5 = 1 at (4, −1).
- 6Write the tangents of slope −1 to x²/16 + y²/9 = 1.
- 7Are the tangents from (5, 12) to x²/100 + y²/69 = 1 perpendicular?
- 8y = x + c touches x²/25 + y²/9 = 1. Find c.
Show answers (try them yourself first)
- 1(6 cos 90°, 5 sin 90°) = (0, 5).
- 2x² + y² = 81.
- 3x² + y² = 9 + 16 = 25.
- 4(4 cos 60°, 6 sin 60°) = (2, 3√3).
- 54x/20 − y/5 = 1, i.e. x − y − 5 = 0.
- 6c² = 16 × 1 + 9 = 25, c = ±5. The lines are x + y − 5 = 0 and x + y + 5 = 0.
- 7Yes. 5² + 12² = 169 and 100 + 69 = 169, so the point is on the director circle.
- 8c² = 25 × 1 + 9 = 34, so c = ±√34.
Hyperbola: standard form
The side of a cooling tower at a power station curves like part of a hyperbola. It is a shape with two separate branches. Building on classes 9–10 : In classes 9–10 we learnt Pythagoras; c² = a² + b² in the hyperbola is the same idea, and only the sign changes compared with the ellipse.
The locus of points for which the absolute value of the difference of the distances from two fixed points (the foci) is a constant 2a is a hyperbola. The constant 2a is less than the distance between the foci. The transverse axis is 2a and the conjugate axis is 2b.
Rule 1: Standard equation — x²/a² − y²/b² = 1. b² = a²(e² − 1), i.e. c² = a² + b² and e = c/a, e > 1. The foci are (±ae, 0), the directrices x = ±a/e and the vertices (±a, 0). If the transverse axis is along the Y-axis the equation is y²/a² − x²/b² = 1 and the foci are (0, ±ae).
In the figure : The difference of the distances from P to the two foci is 2a.
Rule 2: Latus rectum, conjugate and rectangular hyperbola — The length of the latus rectum is 2b²/a. The conjugate hyperbola is y²/b² − x²/a² = 1. When a = b it is a rectangular hyperbola x² − y² = a² and e = √2.
How is the formula obtained? : The foci are (±c, 0) and |PF₁ − PF₂| = 2a. So √((x + c)² + y²) − √((x − c)² + y²) = ±2a. Squaring and simplifying as for the ellipse gives (c² − a²)x² − a²y² = a²(c² − a²). Putting b² = c² − a² gives b²x² − a²y² = a²b², so x²/a² − y²/b² = 1.
- Easy : x²/144 − y²/25 = 1: a = 12, b = 5, c² = 144 + 25 = 169, c = 13. The foci are (±13, 0) and e = 13/12.
- Medium : a = 4, e = 3/2: b² = a²(e² − 1) = 16 × (9/4 − 1) = 20. The equation is x²/16 − y²/20 = 1.
- Exam-type : 9x² − 4y² = 36: divide by 36, x²/4 − y²/9 = 1. a = 2, b = 3, c² = 13, e = √13/2. Latus rectum = 2b²/a = 18/2 = 9.
In a hyperbola the variable of the positive term decides the transverse axis: positive x² means the X-axis, positive y² means the Y-axis. a² is the denominator of the positive term and need not be the larger one.
Using c² = a² − b² as in the ellipse. In the hyperbola c² = a² + b² and e > 1.
Add the denominators of the positive and the negative term to get c². For 9 and 16: c² = 25, c = 5; e = c ÷ a = 5/3. Memory clue : "ellipse subtracts, hyperbola adds".
CET gives it with coefficients like 16x² − 9y² = 144; divide by 144 to get x²/9 − y²/16 = 1 and then a = 3 (not 4).
CET-style example : For the hyperbola 16x² − 9y² = 144 the difference of the distances of any point from the two foci is? (A) 3 (B) 4 (C) 6 (D) 8
Answer : (C). Divide by 144: x²/9 − y²/16 = 1, a = 3. The difference = 2a = 6. Time-saving method : The denominator of the positive term is 9, its root is 3, doubled: 6.
Spot the difference
| Point | Ellipse | Hyperbola |
|---|---|---|
| Equation | x²/a² + y²/b² = 1 | x²/a² − y²/b² = 1 |
| c² | a² − b² | a² + b² |
| Eccentricity | e < 1 | e > 1 |
| Focal distances | sum is constant | difference is constant |
Practice questions
- 1Find the lengths of the transverse and conjugate axes of x²/25 − y²/144 = 1.
- 2Find the foci of x²/64 − y²/36 = 1.
- 3Find the eccentricity of x²/49 − y²/576 = 1.
- 4Write the equation of the hyperbola with vertices (±8, 0) and foci (±17, 0).
- 5Find the length of the latus rectum of x²/36 − y²/64 = 1.
- 6Find the eccentricity of 16x² − 9y² = 144.
- 7Write the equation of the rectangular hyperbola (x² − y² = a²) passing through (5, 4).
- 8Write the directrices of x²/25 − y²/144 = 1.
Show answers (try them yourself first)
- 1The transverse axis is 10 and the conjugate axis is 24.
- 2c² = 64 + 36 = 100, c = 10. The foci are (±10, 0).
- 3c² = 49 + 576 = 625, c = 25. e = c/a = 25/7.
- 4a = 8, c = 17, b² = 289 − 64 = 225. The equation is x²/64 − y²/225 = 1.
- 52b²/a = 2 × 64/6 = 64/3.
- 6x²/9 − y²/16 = 1; c² = 25, e = 5/3.
- 725 − 16 = 9, so a² = 9 and the equation is x² − y² = 9.
- 8c² = 25 + 144 = 169, c = 13, e = 13/5. The directrices are x = ±a/e = ±25/13.
Hyperbola: parametric form, tangents and asymptotes
Two straight railway tracks going far away seem to meet at the horizon but never actually meet. The branches of a hyperbola keep getting closer to two lines in the same way; those lines are the asymptotes. Building on classes 9–10 : In classes 9–10 we learnt sec θ and tan θ and 1 + tan²θ = sec²θ; the parametric form of the hyperbola comes from it.
A point on the hyperbola can be written as (a sec θ, b tan θ). The two lines that the branches approach indefinitely without ever meeting them are the asymptotes.
Rule 1: Tangents — At (x₁, y₁): xx₁/a² − yy₁/b² = 1. At θ: x sec θ/a − y tan θ/b = 1. For slope m: y = mx ± √(a²m² − b²); condition c² = a²m² − b² (a²m² > b²).
Rule 2: Asymptotes and director circle — Asymptotes: y = ±(b/a)x; the combined equation of both lines is x²/a² − y²/b² = 0 (0 in place of 1). Director circle: x² + y² = a² − b² (only when a > b).
In the figure : The diagonals of the green rectangle are the asymptotes; the tangent at P is orange.
How is the formula obtained? : From x²/a² − y²/b² = 1 we get y = ±(b/a)√(x² − a²). When x is very large a² is negligible beside x², so y ≈ ±(b/a)x. For tangency put y = mx + c: (b² − a²m²)x² − 2a²mcx − a²(c² + b²) = 0; making the discriminant zero gives b²c² + b⁴ − a²m²b² = 0, so c² = a²m² − b².
- Easy : x²/49 − y²/16 = 1: a = 7, b = 4, the asymptotes are y = ±(b/a)x = ±4x/7.
- Medium : Tangent to x² − y² = 16 at (5, 3): 5x/16 − 3y/16 = 1, so 5x − 3y − 16 = 0. (Check: 25 − 9 = 16 ✓.)
- Exam-type : Tangents of slope 3/2 to x²/16 − y²/20 = 1: c² = a²m² − b² = 16 × 9/4 − 20 = 16, c = ±4. The lines are y = 3x/2 ± 4, i.e. 3x − 2y ± 8 = 0.
Asymptotes pass through the origin and never meet the hyperbola. In a rectangular hyperbola (a = b) they are y = ±x and perpendicular to each other.
Writing the asymptotes as y = ±(a/b)x. The slope is b/a (root of the negative term ÷ root of the positive term).
Get the asymptotes at once: put 0 in place of 1 in x²/a² − y²/b² = 1. 4x² − 9y² = 36 → 4x² − 9y² = 0. Memory clue : "remove the one, put zero".
In CET the asymptotes and the hyperbola differ only in the constant term; among options with 36 and 0 choose the one with 0.
CET-style example : The combined equation of the asymptotes of 4x² − 9y² = 36 is? (A) 4x² − 9y² = 0 (B) 4x² − 9y² = 1 (C) 9x² − 4y² = 0 (D) 4x² + 9y² = 0
Answer : (A). Replace the constant 36 by 0: 4x² − 9y² = 0. Time-saving method : Keep the left side, write 0 on the right.
Spot the difference
| Point | Ellipse condition | Hyperbola condition |
|---|---|---|
| Condition | c² = a²m² + b² | c² = a²m² − b² |
| Director circle | x² + y² = a² + b² | x² + y² = a² − b² |
| Special lines | none | asymptotes y = ±(b/a)x |
Practice questions
- 1Write the point with θ = 0° on x²/25 − y²/9 = 1.
- 2Write the asymptotes of x²/25 − y²/9 = 1.
- 3Write the tangent to x²/36 − y²/16 = 1 at the vertex (6, 0).
- 4Find the point with θ = 45° on x²/4 − y²/9 = 1.
- 5Write the tangent to x²/9 − y²/16 = 1 at (−5, 16/3).
- 6y = 3x + c touches x²/4 − y²/27 = 1. Find the values of c.
- 7Write the director circle of x²/49 − y²/13 = 1.
- 8Write the equation of the hyperbola with asymptotes y = ±(5/12)x and vertex (12, 0).
Show answers (try them yourself first)
- 1(5 sec 0°, 3 tan 0°) = (5, 0).
- 2b/a = 3/5, so y = ±3x/5.
- 3x = 6, i.e. x − 6 = 0.
- 4(2 sec 45°, 3 tan 45°) = (2√2, 3).
- 5−5x/9 − (16/3)y/16 = 1, i.e. −5x/9 − y/3 = 1. Multiply by −9: 5x + 3y + 9 = 0.
- 6c² = 4 × 9 − 27 = 9, so c = ±3.
- 7x² + y² = 49 − 13 = 36.
- 8a = 12, b/a = 5/12 so b = 5. The equation is x²/144 − y²/25 = 1.
Lesson 7: Conic Sections — quick revision
| Conic | Equation and main formulas |
|---|---|
| Parabola | y² = 4ax; focus (a, 0); directrix x = −a; latus rectum 4a; distance of a point from the focus x₁ + a |
| Tangent to a parabola | yy₁ = 2a(x + x₁); ty = x + at²; condition c = a/m |
| Ellipse | x²/a² + y²/b² = 1; c² = a² − b²; e = c/a < 1; latus rectum 2b²/a; PF₁ + PF₂ = 2a |
| Tangent to an ellipse | xx₁/a² + yy₁/b² = 1; condition c² = a²m² + b² |
| Auxiliary and director circles (ellipse) | x² + y² = a²; x² + y² = a² + b² |
| Hyperbola | x²/a² − y²/b² = 1; c² = a² + b²; e > 1; latus rectum 2b²/a; difference 2a |
| Tangent to a hyperbola | xx₁/a² − yy₁/b² = 1; condition c² = a²m² − b² |
| Asymptotes | y = ±(b/a)x; combined equation x²/a² − y²/b² = 0 |
| Director circle (hyperbola) | x² + y² = a² − b² |
| Parametric form | parabola (at², 2at); ellipse (a cos θ, b sin θ); hyperbola (a sec θ, b tan θ) |
🎯 Most important points for the exam
- In y² = 4ax, coefficient ÷ 4 = a.
- The ellipse subtracts (c² = a² − b²), the hyperbola adds (c² = a² + b²).
- The ellipse condition has + b², the hyperbola condition has − b².
- To write a tangent: x² → xx₁, y² → yy₁, 2x → x + x₁.
- Asymptotes: put 0 in place of 1.
- Director circle: ellipse a² + b², hyperbola a² − b².
First bring the equation to the form with 1 on the right, identify a² and b², then use the formula. After getting the answer, check by putting the given point in the equation.