Conic Section | ISC Class 11 Maths Notes

This note covers sections of a cone, degenerate conics, focus and directrix, eccentricity, standard equations of parabolas, ellipses and hyperbolas, axes, vertices, focal properties, latus rectum, rough sketches and applications.

How does a plane cutting a cone produce different conic sections?

A conic section is the curve formed where a plane intersects a double-napped right circular cone. A plane is a flat surface extending indefinitely. A double-napped cone has two parts, called nappes, meeting at a common point called its vertex.

To form the cone, rotate a straight line around a fixed intersecting vertical line while keeping their angle constant. The fixed line is the cone's axis; the rotating line is a generator. The cone extends indefinitely in both directions.

How do the cutting angles determine the curve?

Let α, read alpha, be the angle between a generator and the cone's axis. Let β, read beta, be the angle between the cutting plane and that axis. The symbol ° denotes degrees. First consider a plane that does not pass through the vertex.

A circle consists of points at a fixed distance from a fixed point, its centre. The distance is its radius. The ellipse, parabola and hyperbola can also be defined by distances, which makes it possible to obtain their equations.

Blue cutting planes cross the black double cones. The circle and ellipse appear on one nappe, the parabolic section extends along one nappe, and the hyperbolic section meets both nappes.

See Figs. 10.4 to 10.7 in your NCERT textbook

What happens when the plane passes through the vertex?

A degenerate conic is a limiting section that becomes a point or straight line arrangement. Passing through the vertex changes the result even when the angle condition resembles that for an ordinary conic.

Each blue plane passes through the cone's vertex. The drawings show a point section, a plane containing one generator, and planes containing two intersecting generators.

See Figs. 10.8 to 10.10 in your NCERT textbook

What do focus, directrix and eccentricity mean?

A locus is the set of points satisfying a specified condition. For a focus-directrix description, choose a fixed point S, called the focus, and a fixed line, called the directrix. The focus does not lie on the directrix for the ordinary conics considered here.

Let P be a variable point on the conic and L the foot of the perpendicular from P to the directrix. Thus PS is the distance from P to S, and PL is the perpendicular distance from P to the directrix.

Definition: The positive constant e in PS = ePL is the eccentricity. Equivalently, e = PS/PL. It compares the distance to a focus with the distance to its corresponding directrix.

For a parabola, e = 1, so the two distances are equal. For a non-circular ellipse, 0 < e < 1; for an ordinary hyperbola, e > 1. These conditions distinguish the three focus-directrix cases.

How will coordinates and distance notation be used?

Write P as P(x, y), where x and y are its signed coordinates along the horizontal x-axis and vertical y-axis. Their intersection is the origin O, with coordinates (0, 0). A superscript ² means the square of a quantity.

The symbol √ denotes the non-negative square root, and | | denotes absolute value, the non-negative magnitude of the enclosed quantity. The symbol ± means that both the positive and negative choices are included.

A chord is a line segment whose endpoints lie on a curve. A latus rectum is a chord through a focus and perpendicular to the conic's relevant axis. Its position and length help connect an equation with a rough sketch.

How is the standard equation of a parabola derived?

Definition: A parabola is the locus of points in a plane equidistant from a fixed point, its focus, and a fixed line, its directrix, with the focus not on that line.

The axis of the parabola is the line through the focus perpendicular to the directrix. The vertex is the point where the parabola meets this axis. It lies halfway between the focus and the directrix along their common perpendicular.

Take the vertex at O(0, 0), the focus at S(a, 0), and the directrix x = −a. Here a > 0 is the distance from the vertex to the focus; the minus sign − indicates a negative coordinate.

Result: The right-opening parabola has equation y² = 4ax

1. Take any point P(x, y) on the parabola. Its perpendicular projection onto the directrix is L(−a, y).
2. By the distance formula, PS = √((x − a)² + y²), while PL = |x + a|.
3. Use PS = PL and square both non-negative distances: (x − a)² + y² = (x + a)².
4. Expand both sides and cancel x² and a². The remaining equation is y² = 4ax.

y² = 4ax therefore follows from the equal-distance condition. Conversely, substituting this equation into PS gives √((x + a)²) = |x + a| = PL, so every point satisfying the equation meets the defining condition.

Since y² cannot be negative and a is positive, x ≥ 0. The curve opens to the right and is symmetric about the x-axis: changing y to −y leaves the equation unchanged. Here symmetry means that reflection in the axis preserves the curve.

The directrix is a vertical blue line to the left of O. The focus F(a, 0) lies on the positive x-axis. A point P(x, y) is joined to F and horizontally to B on the directrix; F and B correspond to S and L above.

See Fig. 10.16 in your NCERT textbook

If the fixed focus lies on the fixed line, the equal-distance locus becomes the line through that point perpendicular to the fixed line. This is the degenerate parabola, rather than the ordinary curved case described by a positive a.

How do the four standard parabolas differ?

Keep a positive in all four standard forms. The squared variable identifies the axis direction, while the sign of the unsquared term identifies the opening direction. Each standard parabola has vertex (0, 0).

The coordinate axes divide the plane into four quadrants. Quadrants I, II, III and IV have coordinate signs (+, +), (−, +), (−, −) and (+, −), respectively. Points on the axes themselves are not inside a quadrant.

Property: The latus rectum of each standard parabola has length 4a

For y² = 4ax, the latus rectum passes through S(a, 0) and is perpendicular to the x-axis. Its line is therefore x = a. Substitution gives y² = 4a², so its endpoints are (a, 2a) and (a, −2a).

The distance between these endpoints is 4a. Each half has length 2a. Reflecting the curve or interchanging the coordinate axes changes the endpoint positions, but leaves the total length unchanged.

The four drawings show right-, left-, upward- and downward-opening curves with vertex O. Each focus is inside its curve, and the blue directrix lies on the opposite side of the vertex.

See Fig. 10.15 in your NCERT textbook

How should a rough sketch be labelled?

Start with the coordinate axes and vertex. Mark the focus, then place the directrix an equal distance from the vertex on the opposite side. Draw the curve opening towards the focus and away from the directrix, keeping its symmetry visible.

For a y² equation, sketch a sideways parabola; for an x² equation, sketch a vertical parabola. Mark the latus rectum through the focus, not through the vertex. Its length describes a chord across the curve, not the vertex-to-focus distance.

How are equations and features of parabolas calculated?

First identify whether the given information is an equation, a focus and directrix, a vertex and focus, or a vertex together with a point on the curve. Use that information to fix both the orientation and the parameter a.

How do you read an equation?

Worked example 1. For y² = 8x, find the focus, axis, directrix and length of the latus rectum.

Answer: Comparing y² = 8x with y² = 4ax gives 4a = 8, so a = 2. The parabola opens right. Its focus is (2, 0), axis is y = 0, directrix is x = −2, and latus rectum has length 8 units.

The coefficient 8 is 4a, not a. Reading the coefficient as the focal distance would place the focus and directrix incorrectly. The same comparison also gives the full latus rectum length directly.

How do you use a vertex and focus?

Worked example 2. Find the equation of the parabola with vertex (0, 0) and focus (0, 2).

Answer: The focus lies above the vertex, so the axis is the y-axis and the curve opens upwards. Use x² = 4ay. The vertex-to-focus distance is a = 2, giving x² = 4(2)y, or x² = 8y.

With a focus and directrix, locate the vertex as their midpoint along the perpendicular from the focus. For the focus (2, 0) and directrix x = −2, this gives vertex (0, 0), a = 2 and equation y² = 8x.

How does a point on the curve determine a?

Worked example 3. A parabola has vertex (0, 0), is symmetric about the y-axis and passes through (2, −3). Find its equation.

Answer: The negative y-coordinate shows that the curve opens downwards. Use x² = −4ay with a positive. Substituting the point gives 4 = −4a(−3) = 12a, hence a = 1/3. Therefore x² = −(4/3)y, or 3x² = −4y.

The vertex condition matters: symmetry about the y-axis alone does not fix where the vertex lies on that axis. After calculating the equation, substitute the given point again to check both the numerical coefficient and its sign.

What defines an ellipse and determines its standard equation?

Definition: An ellipse is the locus of points in a plane whose distances from two fixed points have a constant sum. Those fixed points are the foci, the plural of focus.

The constant sum is greater than the distance between the foci. The midpoint of the foci is the ellipse's centre. The major axis is the segment across the ellipse through the foci; the minor axis passes through the centre perpendicular to it.

The major-axis endpoints are the vertices. For ellipses below, let a be the semi-major axis, half the major-axis length; let b be the semi-minor axis; and let c be the centre-to-focus distance. Thus a > b > 0, and the foci are 2c apart.

Property: The focal sum is 2a and c² = a² − b²

Call the foci S and S′, where the prime on S′ distinguishes the second focus. At a major-axis vertex, the focal distances are a + c and a − c. Their sum is 2a, so PS + PS′ = 2a for every point P on the ellipse.

At a minor-axis endpoint, both focal distances equal √(b² + c²). Their sum must also be 2a. Hence a² = b² + c², or c² = a² − b². The eccentricity is e = c/a, giving b² = a²(1 − e²).

Which standard equation matches each orientation?

The larger denominator lies under the variable along the major axis. If an equation is supplied as x²/a² + y²/b² = 1 with a < b, its major axis is vertical. Relabel its larger semi-axis as the semi-major axis before applying the formulae above.

Two closed curves are centred at O. The first is elongated horizontally with both foci on the x-axis; the second is elongated vertically, with marked endpoints (0, ±a), (±b, 0) and foci (0, ±c).

See Fig. 10.24 in your NCERT textbook

Both standard ellipses are symmetric about both coordinate axes. For the horizontal form, −a ≤ x ≤ a and −b ≤ y ≤ b. These bounds locate the whole curve, not merely its four axis endpoints.

How are an ellipse's directrices and latus rectum found?

A non-circular ellipse has a corresponding directrix for each focus. For the horizontal standard form, the focus (ae, 0) corresponds to x = a/e, and (−ae, 0) corresponds to x = −a/e. Recall that c = ae.

For the vertical form, the directrices are y = ±a/e, with matching focus signs. Since 0 < e < 1, the directrices lie beyond the vertices. The focus-directrix description uses a focus and its matching line, not an arbitrary pairing.

How does the distance ratio recover the ellipse equation?

For S(ae, 0) and directrix x = a/e, apply PS = ePL at P(x, y). Squaring gives (x − ae)² + y² = e²(x − a/e)². Expansion cancels the terms involving the first power of x.

The result is (1 − e²)x² + y² = a²(1 − e²). Using b² = a²(1 − e²) and dividing produces x²/a² + y²/b² = 1. This links the ratio definition to the standard equation.

Result: The latus rectum of an ellipse has length 2b²/a

For the horizontal ellipse, the latus rectum through the right-hand focus lies on x = c = ae. Substitution gives y²/b² = 1 − e² = b²/a². Thus its endpoints have y-coordinates ±b²/a, making the full length 2b²/a.

Each latus rectum is perpendicular to the major axis. The vertical ellipse has the same length formula, with its latus recta horizontal. Here latus recta is the plural of latus rectum.

Worked example 4. Find the foci, vertices, axis lengths, eccentricity and latus rectum length of x²/25 + y²/9 = 1.

Answer: The major axis is horizontal. We have a = 5, b = 3 and c = √(25 − 9) = 4. Foci are (±4, 0), vertices are (±5, 0), and major and minor axis lengths are 10 and 6 units.

The eccentricity is e = 4/5. Each latus rectum has length 2b²/a = 2(9)/5 = 18/5 units. The distances a and b are half-axis lengths, so they must be doubled when stating full axis lengths.

Worked example 5. Find the foci, vertices, axis lengths and eccentricity of 9x² + 4y² = 36.

Answer: Divide by 36 to get x²/4 + y²/9 = 1. The major axis is vertical, with a = 3, b = 2 and c = √5. Foci are (0, ±√5), and vertices are (0, ±3).

The major axis has length 6 units, the minor axis has length 4 units, and e = √5/3. Dividing to make the right-hand side equal to 1 must come before reading the denominators.

How can sufficient information determine an ellipse?

To find a standard ellipse, establish the centre and major-axis direction first. Then determine two independent lengths among a, b and c, using c² = a² − b² to obtain the third. The equation uses the squares of the semi-axis lengths.

How do vertices and foci determine the equation?

Worked example 6. Find the equation of the ellipse with vertices (±13, 0) and foci (±5, 0).

Answer: Their common midpoint is the origin, and the major axis is horizontal. The centre-to-vertex distance is a = 13, while the centre-to-focus distance is c = 5. Hence b² = 169 − 25 = 144.

Put a² = 169 and b² = 144 into the horizontal standard form. The required equation is x²/169 + y²/144 = 1.

The signs in (±5, 0) locate the two foci. They do not make c a negative distance: c = 5 is positive. Similarly, the distance between the two vertices is 26, while a is 13.

How do an axis length and foci determine the equation?

Worked example 7. An ellipse has major-axis length 20 units and foci (0, ±5). Find its equation.

Answer: The centre is (0, 0), and the foci establish a vertical major axis. Half the major-axis length gives a = 10. Since c = 5, b² = a² − c² = 100 − 25 = 75.

The larger squared semi-axis belongs under y². Therefore the equation is x²/75 + y²/100 = 1.

What data are needed with a focus and directrix?

A focus and directrix must be accompanied by sufficient information, such as the eccentricity, to fix the ellipse. If the focus is S(ae, 0), the corresponding directrix is x = a/e, and e is specified, use PS = ePL and simplify as above.

For a vertical axis, interchange the roles of x and y. Check that the calculated eccentricity lies between zero and one and that the major-axis endpoints lie farther from the centre than the foci.

What defines a hyperbola and its two standard orientations?

Definition: A hyperbola is the locus of points in a plane for which the difference between the distances to two fixed foci is constant. Take the distance to the farther focus minus the distance to the nearer focus.

The midpoint of the foci is the centre. The transverse axis passes through both foci; the perpendicular line through the centre is the conjugate axis. The hyperbola meets its transverse axis at its two vertices.

For hyperbolas, let a be half the distance between the vertices and c half the distance between the foci. Define the positive quantity b by b² = c² − a². The transverse and conjugate axis lengths are 2a and 2b.

Property: The focal difference is 2a and c² = a² + b²

For foci S and S′, the branch-independent focal property is |PS − PS′| = 2a. The absolute value accommodates both branches, the two separate parts of the hyperbola. Reversing the branch reverses which named focus is farther away.

An ordinary hyperbola has c > a > 0. Its eccentricity is e = c/a > 1, and b² = a²(e² − 1). Unlike an ellipse, a need not be greater than b: a is associated with the transverse axis.

The positive squared term identifies the transverse-axis direction once the right-hand side is 1. The horizontal hyperbola satisfies x ≤ −a or x ≥ a, so no part lies between its vertices' vertical lines. It has no real intersection with the conjugate axis.

The first drawing has branches opening left and right, with foci (±c, 0) beyond vertices (±a, 0). The second has branches opening upwards and downwards, with foci (0, ±c) beyond vertices (0, ±a).

See Fig. 10.29 in your NCERT textbook

For a rough sketch, plot the centre, vertices and foci first. Draw two branches symmetric about both axes, opening along the transverse axis. A hyperbola with a = b is called an equilateral hyperbola.

How are hyperbola properties used in calculations?

The directrices have the same symbolic positions ±a/e along the transverse direction as for an ellipse. Their positions relative to the vertices differ: because e > 1, a/e < a, so they lie between the centre and the corresponding vertices.

How does the focus-directrix condition give the hyperbola?

For focus S(ae, 0) and directrix x = a/e, squaring PS = ePL again gives (x − ae)² + y² = e²(x − a/e)². Now e² − 1 is positive.

Rearranging gives (e² − 1)x² − y² = a²(e² − 1). Since b² = a²(e² − 1), division gives x²/a² − y²/b² = 1. Interchanging coordinates gives the vertical form.

Result: Each hyperbola latus rectum has length 2b²/a

For a horizontal hyperbola, put x = c into its equation. Then y²/b² = c²/a² − 1 = b²/a². Thus y = ±b²/a, and the distance between the endpoints is 2b²/a. For the vertical form, interchange x and y.

Worked example 8. Find the foci, vertices, eccentricity and latus rectum length of x²/9 − y²/16 = 1.

Answer: The positive x² term gives a horizontal transverse axis. We have a = 3, b = 4 and c = √(9 + 16) = 5. The foci are (±5, 0), and the vertices are (±3, 0).

The eccentricity is e = 5/3. Each latus rectum has length 2b²/a = 32/3 units. Although 16 is the larger denominator, the positive term determines the transverse axis.

Worked example 9. Find the foci, vertices, eccentricity and latus rectum length of y² − 16x² = 16.

Answer: Divide by 16 to obtain y²/16 − x²/1 = 1. The transverse axis is vertical. Thus a = 4, b = 1 and c = √17. Foci are (0, ±√17), and vertices are (0, ±4).

Its eccentricity is e = √17/4. Each latus rectum has length 2b²/a = 2/4 = 1/2 unit.

How do a focal distance and latus rectum determine the equation?

Worked example 10. Find the equation of a hyperbola with foci (0, ±12) and latus rectum length 36 units.

Answer: The centre is the origin and the transverse axis is vertical. We have c = 12 and 2b²/a = 36, so b² = 18a. Using c² = a² + b² gives 144 = a² + 18a.

Hence a² + 18a − 144 = 0, or (a + 24)(a − 6) = 0. Reject a = −24 because a is a length. Thus a = 6 and b² = 108.

The equation is y²/36 − x²/108 = 1, equivalently 3y² − x² = 108.

How can a parabolic shape be used in an application problem?

A geometric application becomes a coordinate problem once the vertex, axis and units are fixed. The chosen coordinate direction must match the standard form. Distinguish a distance along the axis from a distance across the parabola.

How is the width of a parabolic mirror calculated?

Worked example 11. A mirror has a parabolic axial cross-section. Its focus is 5 centimetres (cm) from its vertex, and it is 45 cm deep along its axis. Find the full width of its opening perpendicular to the axis.

Answer: Take the vertex as (0, 0) and the axis along the positive x-axis. Measure x and y in centimetres, abbreviated cm. The focal distance is a = 5, giving y² = 4(5)x = 20x.

At the opening, x = 45. Therefore y² = 20(45) = 900, so y = ±30. The two edges lie 30 cm on opposite sides of the axis; the full width is 30 − (−30) = 60 cm.

The drawing places the vertex at O and the mirror's axis along the x-axis. The upper rim is labelled A and the lower rim B. A dashed horizontal arrow marks the depth 45, and AB spans the opening.

See Fig. 10.31 in your NCERT textbook

The calculated positive y-value is a half-width, not the full opening. The negative value represents the opposite side of the same cross-section. Both values are necessary when finding the distance across the opening.

What checks prevent a wrong interpretation?

1. Identify the vertex and choose the coordinate origin there.
2. Align a coordinate axis with the symmetry axis and state the unit of length.
3. Use the focus-to-vertex distance as a in the appropriate standard equation.
4. Substitute the stated depth as the coordinate along the axis, then calculate the full requested distance.

Finally, connect the algebraic answer to the requested geometric quantity. A focus coordinate, an axis length, a latus rectum length and an opening width are different measurements even when the same equation supplies all of them.

Glossary

• Conic section — A curve obtained by intersecting a right circular cone with a plane.
• Nappe — One of the two parts of a double cone separated by its vertex.
• Generator — A straight line whose rotation around the cone's axis produces its surface.
• Locus — The set of all points satisfying a stated geometric condition.
• Focus — A fixed point used in the distance conditions defining a conic.
• Directrix — A fixed line used with a corresponding focus to define a conic.
• Eccentricity — The constant ratio of distance to a focus to perpendicular distance from its corresponding directrix.
• Vertex — An intersection of a parabola, ellipse or hyperbola with its axis, major axis or transverse axis respectively.
• Latus rectum — A chord through a focus perpendicular to the parabola's axis, ellipse's major axis or hyperbola's transverse axis.
• Major axis — The longer axis segment of an ellipse, passing through both its foci.
• Minor axis — The shorter axis segment of an ellipse, perpendicular to its major axis through the centre.
• Transverse axis — The axis of a hyperbola passing through both foci and both vertices.
• Conjugate axis — The line through a hyperbola's centre perpendicular to its transverse axis.
• Degenerate conic — A limiting conic section that becomes a point, straight line or pair of intersecting lines.

Common errors and misconceptions

• Misconception: In y² = 8x, the focal distance is 8. Correct: Compare with y² = 4ax: 4a = 8, so a = 2.
• Misconception: A y² term makes the parabola's axis vertical. Correct: In the standard parabolas, a y² term gives a horizontal axis; an x² term gives a vertical axis.
• Misconception: The major-axis length of an ellipse equals a. Correct: With a denoting the semi-major axis, the full major-axis length is 2a.
• Misconception: Use c² = a² + b² for both ellipses and hyperbolas. Correct: An ellipse uses c² = a² − b²; a hyperbola uses c² = a² + b².
• Misconception: The larger hyperbola denominator identifies the transverse axis. Correct: After writing the right-hand side as 1, identify the variable in the positive squared term.
• Misconception: PS − PS′ = 2a holds with one fixed ordering on both hyperbola branches. Correct: Subtract the nearer distance from the farther, or write |PS − PS′| = 2a.
• Misconception: A latus rectum runs through the vertex. Correct: It passes through a focus and is perpendicular to the relevant axis, with both endpoints on the conic.
• Misconception: The positive y-value in the mirror example is the full opening width. Correct: It gives half the width; the full width is the distance between the positive and negative y-values.

Exam-style questions with model answers

Q1. Define a parabola using a focus and directrix, and state the condition on the position of the focus for an ordinary parabola. [2 marks]

1. A parabola is the set of points in a plane equidistant from a fixed point, called the focus, and a fixed line, called the directrix.
2. The focus must not lie on the directrix; placing it on that line gives a degenerate straight-line locus.

Q2. For the parabola y² = 8x, find its focus, the equation of its directrix and the length of its latus rectum. [3 marks]

1. Compare the equation with y² = 4ax, where a is the vertex-to-focus distance. Since 4a = 8, a = 2, and the focus is (2, 0).
2. The curve opens right, so its directrix lies to the left of the vertex. Its equation is x = −a = −2.
3. The full latus rectum length is 4a = 4(2) = 8 units.

Q3. Find the equation of the parabola with vertex (0, 0) and focus (0, 2). [3 marks]

1. The line through the vertex and focus is the y-axis. Since the focus is above the vertex, the parabola opens upwards.
2. The standard equation is x² = 4ay, where a is the positive vertex-to-focus distance. The given points give a = 2.
3. Substituting gives x² = 4(2)y, so the required equation is x² = 8y.

Q4. For the ellipse 9x² + 4y² = 36, find the direction of the major axis, the foci, the vertices and the eccentricity. [4 marks]

1. Divide by 36: x²/4 + y²/9 = 1. The larger denominator is under y², so the major axis is vertical.
2. Let a and b be the semi-major and semi-minor lengths and c the centre-to-focus distance. Then a = 3, b = 2 and c² = 9 − 4 = 5. The foci are (0, ±√5).
3. The vertices are the major-axis endpoints (0, ±a), giving (0, 3) and (0, −3).
4. The eccentricity is e = c/a = √5/3.

Q5. An ellipse has major-axis length 20 units and foci (0, 5) and (0, −5). Find its equation. [4 marks]

1. The centre is the midpoint of the foci, (0, 0). The major axis passes through them and therefore lies along the y-axis.
2. The semi-major length a is half the full major-axis length: a = 20/2 = 10. The centre-to-focus distance is c = 5.
3. For semi-minor length b, use b² = a² − c² = 100 − 25 = 75.
4. The vertical standard form is x²/b² + y²/a² = 1, giving x²/75 + y²/100 = 1.

Q6. For x²/9 − y²/16 = 1, identify the transverse-axis direction and find the foci, vertices, eccentricity and latus rectum length. [5 marks]

1. The right-hand side is 1 and the positive term contains x². Hence the transverse axis is horizontal. Write a = 3 and b = 4, the transverse and conjugate half-lengths.
2. Let c be the centre-to-focus distance. Using the hyperbola relation gives c = √(a² + b²) = √(9 + 16) = 5. Its foci are (±5, 0).
3. The vertices are (±a, 0), so their coordinates are (3, 0) and (−3, 0).
4. The eccentricity is e = c/a = 5/3, which is greater than one.
5. Each latus rectum has length 2b²/a = 2(16)/3 = 32/3 units.

Q7. A mirror has a parabolic axial cross-section, a focus 5 cm from its vertex and a depth of 45 cm measured along its axis. Find the full opening width perpendicular to that axis. [6 marks]

1. Place the vertex at (0, 0) and take the mirror's axis along the positive x-axis. Measure both coordinates in centimetres.
2. Let a be the vertex-to-focus distance. The stated focal distance gives a = 5, so the focus is (5, 0).
3. The cross-section has equation y² = 4ax = 4(5)x = 20x.
4. The rim is 45 cm along the axis from the vertex, so substitute x = 45 to obtain y² = 900.
5. The upper and lower edges have y-coordinates 30 and −30, respectively.
6. The full width is the difference between these coordinates: 30 − (−30) = 60 cm.

Q8. A hyperbola has foci (0, 12) and (0, −12), and each latus rectum has length 36 units. Find its equation. [6 marks]

1. The midpoint of the foci is the origin. The transverse axis is vertical, and the centre-to-focus distance c is 12.
2. Let a and b be the transverse and conjugate half-lengths. The latus rectum condition gives 2b²/a = 36, hence b² = 18a.
3. Use c² = a² + b² to obtain 144 = a² + 18a, or a² + 18a − 144 = 0.
4. Factorising gives (a + 24)(a − 6) = 0. Since a is a positive length, take a = 6.
5. Substitute into b² = 18a to get b² = 108; also a² = 36.
6. The vertical standard form gives y²/36 − x²/108 = 1, equivalently 3y² − x² = 108.

Key takeaways

• The cutting plane's angle and whether it passes through the cone's vertex determine the type of conic section.
• The focus-directrix condition PS = ePL compares distance to a fixed focus with perpendicular distance to its corresponding directrix.
• For standard parabolas, keep a positive, identify the opening from the equation, and use 4a for the latus rectum length.
• An ellipse has constant focal sum 2a and satisfies c² = a² − b² when a is the semi-major length.
• A hyperbola has constant absolute focal difference 2a and satisfies c² = a² + b².
• The larger denominator identifies an ellipse's major axis; the positive term identifies a hyperbola's transverse axis in standard form.
• Ellipses and hyperbolas have latus rectum length 2b²/a, but their eccentricities and relationships between a, b and c differ.
• In applications, state the coordinate origin, axis direction and length units before substituting data into a conic equation.

Test yourself

What sections can result when the cutting plane passes through the cone's vertex?

A point, a straight line or a pair of intersecting straight lines, depending on the cutting plane's angle.

What does PL represent in PS = ePL, where S is a focus and P lies on the conic?

PL is the perpendicular distance from P to the corresponding directrix; L is the perpendicular's foot on that line.

For a > 0, where does x² = −4ay open, and what are its focus and directrix?

It opens downwards, with focus (0, −a) and directrix y = a. Its axis is the y-axis.

For x²/25 + y²/9 = 1, which axis is the major axis and what is its length?

The major axis is horizontal because 25 is the larger denominator. Its length is twice √25, giving 10 units.

What relation connects a, b and c for an ellipse, with a and b its semi-major and semi-minor lengths and c its centre-to-focus distance?

The relation is c² = a² − b²; equivalently, a² = b² + c².

How does the definition of focal difference apply to both branches of a hyperbola?

Subtract the nearer focal distance from the farther one, or use an absolute value to make the difference positive.

Why does x²/9 − y²/16 = 1 have a horizontal transverse axis despite the larger denominator under y²?

The transverse direction is determined by the positive term, which contains x², rather than by the larger denominator.

In the mirror example, why does y = 30 cm give an opening width of 60 cm?

The two edges have coordinates y = 30 cm and y = −30 cm, so their separation is 60 cm.