Which part of Analytical Geometry II is most important?

Samacheer Kalvi 12th Maths Chapter 5 Exercise Solutions

Chapter 5 — Exercise Solutions Guide

This chapter is formula-heavy. Draw a small rough diagram whenever possible; it reduces mistakes in focus, directrix, tangent, and normal questions.

How to Use This Page

1. Start with Key Concepts to quickly recall definitions and formulas. 2. Attempt MCQs, Fill in the Blanks, and True/False to test basic understanding. 3. Try Match the Following to connect terms with their properties. 4. Use Answer Briefly (2-mark) for short-answer exam practice. 5. Study Answer in Detail (5-mark) for long-answer exam practice — these follow the step-by-step pattern board examiners expect. 6. Review Common Mistakes before every mock test to avoid losing marks on silly errors. 7. Check the Exam Checklist the night before your exam to ensure you have not missed any topic.

Key Concepts (Quick Recap)

Term	Definition
Conic section	Locus of a point whose distance from a fixed point (focus) bears a constant ratio (eccentricity) to its distance from a fixed line (directrix)
Parabola	Conic with eccentricity e = 1; standard form `y² = 4ax`
Ellipse	Conic with eccentricity 0 < e < 1; standard form `x²/a² + y²/b² = 1` where a > b
Hyperbola	Conic with eccentricity e > 1; standard form `x²/a² - y²/b² = 1`
Eccentricity (e)	Ratio of distance from focus to distance from directrix; determines the type of conic
Focus	Fixed point used to define the conic; parabola has 1, ellipse and hyperbola have 2 each
Directrix	Fixed line corresponding to each focus; conic point distance ratio equals e
Latus rectum	Chord through focus perpendicular to axis; length = 4a for parabola, `2b²/a` for ellipse, `2b²/a` for hyperbola
Tangent	Line that touches the conic at exactly one point; slope equals `dy/dx` at that point
Normal	Line perpendicular to the tangent at the point of contact
Vertex	Point where the conic meets its axis; (0, 0) for standard forms
Centre	Midpoint of line joining the two foci; exists for ellipse and hyperbola but not parabola
Asymptote	Line the hyperbola approaches but never touches; equations `y = ±(b/a)x` for standard hyperbola
Condition for tangent (parabola)	Line `y = mx + c` touches `y² = 4ax` if `c = a/m`
Condition for tangent (ellipse)	Line `y = mx + c` touches `x²/a² + y²/b² = 1` if `c² = a²m² + b²`

Formula Recap

Conic	Standard equation
Parabola	y² = 4ax
Ellipse	x²/a² + y²/b² = 1
Hyperbola	x²/a² - y²/b² = 1

Parabola Quick Facts

For y² = 4ax:

Focus = (a, 0)
Directrix = x = -a
Vertex = (0, 0)
Axis = x-axis

Standard Solution Pattern

1. Identify the conic from the equation. 2. Compare with standard form. 3. Find required values: a, b, focus, vertex, eccentricity. 4. Use tangent or normal formula if asked. 5. Substitute carefully and simplify.

Tangent Pattern

For parabola y² = 4ax, tangent at (x1, y1) is:

yy1 = 2a(x + x1)

For ellipse x²/a² + y²/b² = 1, tangent at (x1, y1) is:

xx1/a² + yy1/b² = 1

I. Choose the Best Answer (MCQs)

1. The eccentricity of a parabola is:

(The eccentricity of a parabola is always 1 by definition.)

(a) 0
(b) less than 1
(c) 1
(d) greater than 1

2. For the ellipse x²/25 + y²/9 = 1, the value of c is:

(c² = a² - b² = 25 - 9 = 16, so c = 4.)

(a) 3
(b) 5
(c) 4
(d) 16

3. The focus of the parabola y² = 16x is:

(4a = 16 gives a = 4, so focus = (a, 0) = (4, 0).)

(a) (2, 0)
(b) (8, 0)
(c) (4, 0)
(d) (16, 0)

4. The length of the latus rectum of x²/16 + y²/9 = 1 is:

(Latus rectum = 2b²/a = 2(9)/4 = 18/4 = 9/2.)

(a) 9/4
(b) 32/9
(c) 9/2
(d) 16/9

5. The directrix of y² = 8x is:

(4a = 8 gives a = 2; directrix: x = -a = -2.)

(a) x = 2
(b) x = -2
(c) y = 2
(d) y = -2

6. If the eccentricity of an ellipse is 1/2 and a = 6, then b equals:

(e = c/a = 1/2 so c = 3; b² = a² - c² = 36 - 9 = 27; b = 3√3.)

(a) 3
(b) 3√3
(c) 6
(d) √3

7. The asymptotes of x²/9 - y²/4 = 1 are:

(Asymptotes: y = ±(b/a)x = ±(2/3)x.)

(a) y = ±(2/3)
(b) y = ±(2/3)x
(c) y = ±(3/2)x
(d) x = ±(2/3)y

8. The condition for y = mx + c to be a tangent to y² = 4ax is:

(The condition for tangency to a parabola is c = a/m.)

(a) c = am
(b) c = am²
(c) c = a/m
(d) c = -a/m

9. The centre of the hyperbola x²/16 - y²/9 = 1 is:

(Standard hyperbola has centre at origin.)

(a) (0, 0)
(b) (4, 3)
(c) (4, 0)
(d) (0, 3)

10. For y² = 4ax, the equation of the normal at (x1, y1) is:

(Normal slope = -1/(tangent slope) = -2a/y1.)

(a) yy1 = 2a(x + x1)
(b) y - y1 = (y1/2a)(x - x1)
(c) y - y1 = (-2a/y1)(x - x1)
(d) yy1 + 2a(x + x1) = 0

II. Fill in the Blanks

1. The eccentricity of a circle is ______.

Answer0

2. For the parabola y² = 4ax, the latus rectum length is ______.

Answer4a

3. The foci of the ellipse x²/a² + y²/b² = 1 are ______.

Answer(±c, 0) where c² = a² - b²

4. The relation between a, b, and e for an ellipse is ______.

Answerb² = a²(1 - e²)

5. The equation of the directrix of x²/a² + y²/b² = 1 (a > b) is ______.

Answerx = ±a/e

6. The slope of the tangent to y² = 4ax at (x1, y1) is ______.

Answer2a/y1

7. The number of foci in a hyperbola is ______.

Answer2

8. For the hyperbola x²/a² - y²/b² = 1, the relation e² = 1 + b²/a² gives eccentricity always greater than ______.

Answer1

9. The length of latus rectum of the hyperbola x²/a² - y²/b² = 1 is ______.

Answer2b²/a

10. If e = 1, the conic is a ______.

Answerparabola

III. True or False

1. The eccentricity of an ellipse is always greater than 1. False. Correction: The eccentricity of an ellipse is always between 0 and 1 (0 < e < 1).

2. A parabola has one focus and one directrix. True.

3. The latus rectum of the ellipse x²/25 + y²/16 = 1 is 32/5. True. (2b²/a = 2(16)/5 = 32/5.)

4. The centre of a parabola lies on the directrix. False. Correction: A parabola does not have a centre; only ellipse and hyperbola have centres.

5. Every diameter of an ellipse passes through its centre. True.

6. The asymptotes of a hyperbola are perpendicular to each other when a = b. True. (In a rectangular hyperbola, asymptotes are perpendicular.)

7. The tangent to a parabola at the vertex is parallel to the directrix. False. Correction: The tangent at the vertex is parallel to the latus rectum (i.e., perpendicular to the axis), not the directrix.

8. For y² = 4ax, the point (at², 2at) lies on the parabola for all real t. True. (Substituting: (2at)² = 4a(at²) gives 4a²t² = 4a²t², which holds.)

IV. Match the Following

Column A	Column B
1. Eccentricity of parabola	(d) e = 1
2. Eccentricity of ellipse	(f) 0 < e < 1
3. Eccentricity of hyperbola	(a) e > 1
4. Latus rectum of `y² = 4ax`	(g) 4a
5. Latus rectum of ellipse	(h) 2b²/a
6. Focus of `y² = 4ax`	(b) (a, 0)
7. Centre of ellipse	(e) (0, 0)
8. Asymptotes of `x²/a² - y²/b² = 1`	(c) y = ±(b/a)x

V. Answer Briefly (2-mark)

1. Find the focus and directrix of the parabola y² = 20x.

Compare with y² = 4ax:  4a = 20, so a = 5
Focus = (a, 0) = (5, 0)
Directrix: x = -a = -5

The focus is (5, 0) and the directrix is x = -5.

2. Find the eccentricity and foci of the ellipse x²/36 + y²/16 = 1.

a² = 36, b² = 16, so a = 6, b = 4
c² = a² - b² = 36 - 16 = 20, so c = 2√5
e = c/a = 2√5/6 = √5/3
Foci = (±2√5, 0)

The eccentricity is √5/3 and the foci are (±2√5, 0).

3. Find the equation of the tangent to the parabola y² = 8x at (2, 4).

y² = 8x → 4a = 8, so a = 2
Tangent at (x1, y1) = (2, 4):
yy1 = 2a(x + x1)
y(4) = 2(2)(x + 2)
4y = 4x + 8
y = x + 2

The tangent equation is y = x + 2.

4. Identify the conic: x²/9 - y²/16 = 1. Find its eccentricity.

Since the equation has a minus sign between x² and y² terms,
it is a hyperbola.
a² = 9, b² = 16 → a = 3, b = 4
e² = 1 + b²/a² = 1 + 16/9 = 25/9
e = 5/3

The conic is a hyperbola with eccentricity 5/3.

5. Find the vertex and axis of the parabola x² = 12y.

Compare with x² = 4ay:  4a = 12, so a = 3
Vertex = (0, 0), Axis = y-axis (vertical)

The vertex is (0, 0) and the axis is the y-axis.

VI. Answer in Detail (5-mark)

1. Derive the standard equation of the ellipse with foci (±c, 0) and major axis 2a. State the relation between a, b, and c.

Let P(x, y) be any point on the ellipse. By definition, the sum of distances from P to the two foci equals 2a.

PF1 + PF2 = 2a
√[(x + c)² + y²] + √[(x - c)² + y²] = 2a

Move one radical to the right side and square both sides:

√[(x + c)² + y²] = 2a - √[(x - c)² + y²]
(x + c)² + y² = 4a² - 4a√[(x - c)² + y²] + (x - c)² + y²

Simplify and isolate the radical:

4cx = 4a² - 4a√[(x - c)² + y²]
√[(x - c)² + y²] = a - cx/a

Square again:

(x - c)² + y² = a² - 2cx + c²x²/a²
x² - 2cx + c² + y² = a² - 2cx + c²x²/a²

Cancel -2cx on both sides and rearrange:

x²(1 - c²/a²) + y² = a² - c²
x²(a² - c²)/a² + y² = a² - c²

Divide throughout by a²(a² - c²):

x²/a² + y²/(a² - c²) = 1

Substituting b² = a² - c², we get x²/a² + y²/b² = 1. The relation is c² = a² - b².

2. Find the equation of the tangent and normal to the ellipse x²/25 + y²/9 = 1 at the point (4, 3/5 * √(9)). Simplify to general form.

First, verify the point lies on the ellipse:

Check: (4)²/25 + (3/5 * 3)²/9 = 16/25 + (9/5)²/9
Wait — let us take point (4, 9/5) on the ellipse.
16/25 + (81/25)/9 = 16/25 + 81/225 = 16/25 + 9/25 = 25/25 = 1 ✓

Tangent at (x1, y1) = (4, 9/5):

xx1/a² + yy1/b² = 1
x(4)/25 + y(9/5)/9 = 1
4x/25 + 9y/45 = 1
4x/25 + y/5 = 1

Multiply by 25:

4x + 5y = 25

Normal slope = -1/(tangent slope). From tangent: slope = -4/5. Normal slope = 5/4.

Normal: y - 9/5 = (5/4)(x - 4)
5(y - 9/5) = 4(x - 4) ... wait let me redo.
y - y1 = m(x - x1)
y - 9/5 = (5/4)(x - 4)
4y - 36/5 = 5x - 20
5x - 4y = 20 - 36/5 = 64/5
25x - 20y = 64

The tangent is 4x + 5y = 25 and the normal is 25x - 20y = 64.

3. A line y = mx + c is a tangent to the hyperbola x²/a² - y²/b² = 1. Derive the condition for tangency and state it.

Substitute y = mx + c into the hyperbola equation:

x²/a² - (mx + c)²/b² = 1
x²/a² - (m²x² + 2mcx + c²)/b² = 1

Multiply by a²b²:

b²x² - a²(m²x² + 2mcx + c²) = a²b²
(b² - a²m²)x² - 2a²mcx - (a²c² + a²b²) = 0

For the line to be a tangent, this quadratic in x must have exactly one solution. So discriminant = 0:

(2a²mc)² - 4(b² - a²m²)(-a²c² - a²b²) = 0
4a⁴m²c² + 4(b² - a²m²)(a²c² + a²b²) = 0
a⁴m²c² + (b² - a²m²)(a²c² + a²b²) = 0
a⁴m²c² + a²b²c² + a²b⁴ - a⁴m²c² - a⁴b²m² = 0
a²b²c² + a²b⁴ - a⁴b²m² = 0
b²c² + b⁴ - a²b²m² = 0
c² + b² - a²m² = 0
c² = a²m² - b²

The condition for tangency to a hyperbola is c² = a²m² - b².

4. Find the centre, vertices, foci, eccentricity, and latus rectum of the ellipse 9x² + 25y² = 225.

Divide by 225:

x²/25 + y²/9 = 1
a² = 25, b² = 9, so a = 5, b = 3
c² = a² - b² = 25 - 9 = 16, c = 4

Results:

Centre = (0, 0)
Vertices = (±5, 0)
Foci = (±4, 0)
Eccentricity e = c/a = 4/5
Latus rectum = 2b²/a = 18/5

5. Find the equation of the tangent and normal to the parabola y² = 12x at the point (3, 6).

y² = 12x → 4a = 12, a = 3
Point (x1, y1) = (3, 6): Check y1² = 36 = 12(3) = 36 ✓

Tangent: yy1 = 2a(x + x1)

y(6) = 2(3)(x + 3)
6y = 6(x + 3)
6y = 6x + 18
y = x + 3

Normal slope = -1/(tangent slope) = -1/1 = -1

y - 6 = -1(x - 3)
y - 6 = -x + 3
y = -x + 9  or  x + y = 9

The tangent is y = x + 3 and the normal is x + y = 9.

Worked Examples

Example 1 — Centre and radius of a circle

Find the centre and radius of x² + y² − 4x − 6y − 12 = 0.

Compare with x² + y² + 2gx + 2fy + c = 0:
2g = −4 → g = −2,  2f = −6 → f = −3,  c = −12
Centre = (−g, −f) = (2, 3)
Radius = √(g² + f² − c) = √(4 + 9 + 12) = √25 = 5

Example 2 — Parabola

Find the focus and directrix of y² = 12x.

Compare with y² = 4ax:  4a = 12 → a = 3
Focus = (a, 0) = (3, 0)
Directrix: x = −a → x = −3

Example 3 — Ellipse

For x²/25 + y²/9 = 1, find the foci and eccentricity.

a² = 25, b² = 9  →  a = 5, b = 3
c = √(a² − b²) = √(25 − 9) = 4
Foci = (±4, 0),  eccentricity e = c/a = 4/5

Common Mistakes to Avoid

❌ Confusing ellipse sign (+) with hyperbola sign (−) in the standard equation → ✅ Ellipse: x²/a² + y²/b² = 1; Hyperbola: x²/a² − y²/b² = 1
❌ Using wrong tangent formula for each conic → ✅ Memorize the correct form: parabola yy1 = 2a(x+x1), ellipse xx1/a² + yy1/b² = 1, hyperbola xx1/a² − yy1/b² = 1
❌ Forgetting to compare with standard form before extracting a and b → ✅ Always rewrite the equation in standard form first; then identify a² and b²
❌ Computing c² = a² − b² for hyperbola → ✅ For ellipse: c² = a² − b²; for hyperbola: c² = a² + b²
❌ Writing latus rectum as 2a²/b instead of 2b²/a → ✅ Latus rectum of ellipse/hyperbola = 2b²/a (always divide b² by a)
❌ Mixing up y² = 4ax (right-opening) with x² = 4ay (upward-opening) → ✅ Check which variable is squared: y² means horizontal axis, x² means vertical axis
❌ Not simplifying the final tangent or normal equation → ✅ Always simplify to general form Ax + By + C = 0
❌ Using a instead of c for focus in ellipse → ✅ Focus is at (±c, 0) where c² = a² − b², not at (±a, 0)

Exam Checklist

Standard form is identified (rewrite the equation first if needed)
Values of a and b are extracted correctly (remember a² and b², not a and b directly)
Correct sign pattern: ellipse (+), hyperbola (−)
Eccentricity formula used correctly for each conic type
Focus, directrix, and vertex computed with correct formulas
Tangent formula matches the conic (parabola vs ellipse vs hyperbola)
Normal slope is -1/(tangent slope), not the same as tangent slope
Latus rectum formula applied: 4a (parabola), 2b²/a (ellipse, hyperbola)
Condition for tangency derived or recalled correctly (c = a/m for parabola, c² = a²m² + b² for ellipse)
Final equation is simplified and written in the required form (general or slope form)
Diagrams drawn for conic problems to verify answers visually

Frequently Asked Questions

Standard equations of parabola, ellipse and hyperbola, tangent and normal formulas, and identifying conic parameters are very important.