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8 62

43

Vol. XXXIV

No. 11

31

55

November 2016

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70 81

CONTENTS

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22

Class XI

8 Concept Boosters 22 Brain @ Work 31 Ace Your Way (Series 7) 41 MPP-5

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Class XII

55 Ace Your Way (Series 7) 62 Challenging Problems

Mathematics Today Chemistry Today Physics For You Biology Today

66 MPP-5 69 Maths Musing Problem Set - 167 70 Mock Test Paper - WB JEE 77 JEE Work Outs 81 Math Archives 83 Maths Musing Solutions 84 You Ask We Answer

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MATHEMATICS TODAY | NOVEMBER‘16

7

This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging. *ALOK KUMAR, B.Tech, IIT Kanpur

MEASURE OF ANGLES There are three system for measuring angles 1. Sexagesimal or English system : 1 right angle = 90 degree(= 90°) 1° = 60 minutes (= 60′) 1′ = 60 seconds (= 60′′) 2. Centesimal or French system : 1 right angle = 100 grades (= 100g) 1 grade = 100 minutes (= 100′) 1 minute = 100 seconds (= 100′′) 3. Circular system : The measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle. 1 radian = 57°17′44.8′′ ≈ 57°17′45′′.   If s is the length of an arc of a circle of radius r, then the angle θ (in radians) subtended by this arc at the centre of the circle is given by s or s = rθ r i.e., Length of arc = radius × angle in radians θ=





 

 

c

Trigonometrical Function sinx cosx tanx

Domain

Range

R R

[–1, 1] [–1, 1] R

π ⎧ ⎫ R − ⎨(2n + 1) , n ∈ I ⎬ 2 ⎩ ⎭

cosecx secx

R – {nπ, n ∈ I}

cotx

R – {nπ, n ∈ I}

R – {x : –1 < x < 1} π ⎧ ⎫ R–{x : –1 sin1° Since, value of sinθ is increasing 2. (d) : We have, sinθ + cosecθ = 2 ⇒ sin2θ + 1 = 2sinθ ⇒ sin2θ – 2sinθ + 1 = 0 ⇒ (sinθ – 1)2 = 0 ⇒ sinθ = 1 1 ⇒ sin10θ + cosec10θ = (1)10 + 10 = 2 (1) 4 3. (b) : Given that sin θ = − and θ lies in the III 5 quadrant. ⇒ cos θ = 1 −

(θ ∈ R, A, B are constants)

16

31 54. In a ΔABC, a = 5, b = 4 and cos ( A − B ) = , then 32 c must be

3 (1 − m2 ) t 2 + (4m2 + 6m − 4) t − 8m = 0 ⇒ (3t − 4)[(1 − m2 ) t + 2m] = 0 , 2m 4 which is true if t = tan θ = or tan θ = 2 3 m −1 ⇒

5. (b) : We have sinx + siny = 3(cosy – cosx) ⇒ sinx + 3cosx = 3cosy – siny ...(i) ⇒ rcos(x – α) = rcos(y + α), 1 where r = 10 , tan α = 3 ⇒ x – α = ±(y + α) ⇒ x = –y or x – y = 2α Clearly, x = –y satisfies (i) sin 3x − sin 3 y ∴ = = −1 sin 3 y sin 3 y 6. (b) : We have, tanθ + secθ = ex ∴ secθ – tanθ = e–x From (i) and (ii), we get 2 2 sec θ = e x + e − x ⇒ cos θ = x − x . e +e

...(i) ...(ii)

1 = 2 cos α x 1 ⇒ x2 + = 4 cos2 α − 2 2 x 1 ⇒ x2 + = 2(2 cos2 α − 1) = 2 cos 2α x2 1 Similarly, x n + n = 2 cos nα x A A 2A 1 + sin A − cos A 2 sin 2 + 2 sin 2 cos 2 8. (c) : = 1 + sin A + cos A 2 cos2 A + 2 sin A cos A 2 2 2 A⎛ A A⎞ 2 sin ⎜ sin + cos ⎟ A 2⎝ 2 2⎠ = tan = A⎛ A A⎞ 2 2 cos ⎜ cos + sin ⎟ ⎠ ⎝ 2 2 2 7. (d) : We have, x +

sin φ

x sin θ = . sin(θ + φ) y sin φ 12. (d) : (m + n) = 2tanθ, (m – n) = 2sinθ ∴ m2 – n2 = 4tanθ · sinθ Similarly, y =

2

;∴

2

and 4 mn = 4 tan θ − sin θ = 4 sin θ ⋅ tan θ

...(i) ...(ii)

=

From (i) and (ii), m2 − n2 = 4 mn 13. (a) : As given, 1 + tan θ = m ⇒ 1 + tan2 θ = m tan θ tan θ ⇒ sec2θ = mtanθ ...(i) and secθ – cosθ = n ⇒ sec2θ – 1 = nsecθ ⇒ tan2θ = nsecθ ⇒ tan4θ = n2sec2θ = n2 m tanθ {by (i)} ⇒ tan3θ = n2m [∵ tanθ ≠ 0] ⇒ tanθ = (n2m)1/3 ...(ii) Also, sec2θ = mtanθ = m(n2m)1/3 {by (i) and (ii)} ⇒ m(mn2)1/3 – (n2m)2/3 = 1 ⇒ m(mn2)1/3 – n(nm2)1/3 = 1 14. (c) : We have, sinx + sin2x = 1 or sinx = 1 – sin2 x or sinx = cos2x ∴ cos12x + 3cos10x + 3cos8x + cos6x – 2 = sin6x + 3sin5x + 3sin4 x + sin3x – 2 = (sin2 x)3 + 3(sin2 x)2 sinx + 3(sin2x)(sinx)2 + (sinx)3 – 2 2 3 3 = (sin x + sinx) – 2 = (1) – 2 = – 1. 15. (c) : We have, xsin3α + ycos3α = sinαcosα ... (i) and xsinα – ycosα = 0 ... (ii) From (i) and (ii), we get ⇒ ycosα sin2α + ycos3α = sinα cosα ⇒ ycosα{sin2α + cos2α} = sinα cosα ⇒ ycosα = sinα cosα ⇒ y = sinα and x = cosα Hence, x2 + y2 = sin2α + cos2α = 1 16. (d) : Since, sin2θ ≤ 1

Adding (i) and (ii) we get, xy + 1 + (x – y) = 0 y −1 ⇒ x= y +1 x sin φ 11. (b) : We have tan θ = 1 − x cos φ ⇒ xsinφ = tanθ – xcosφ tanθ tan θ ⇒ x= sin φ + cos φ tan θ sin θ sin θ = = cos θ sin φ + cos φ sin θ sin(θ + φ)

x2 + y2 + 1 ≤1 2x x2 + y2 – 2x + 1 ≤ 0 (x – 1)2 + y2 ≤ 0 It is possible, iff x = 1 and y = 0, i.e., It also depends on value of y. 17. (d) : Given that tanθ – cotθ = a and sinθ + cosθ = b Now, (b2 – 1)2 (a2 + 4) = {(sinθ + cosθ)2 – 1}2 {(tanθ – cotθ)2 + 4} = [1 + sin2θ – 1]2 [tan2θ + cot2θ – 2 + 4] = sin22θ(cosec2θ + sec2θ) 1 ⎤ ⎡ 1 = 4 sin2 θ cos2 θ ⎢ 2 + ⎥=4 ⎣ sin θ cos2 θ ⎦

...(i) ...(ii)

MATHEMATICS TODAY | NOVEMBER‘16

17

9. (a) : Given that 2ycosθ = xsinθ ...(i) and 2xsecθ – ycosecθ = 3 ...(ii) 2x y ⇒ − =3 cos θ sin θ ⇒ 2xsinθ – ycosθ – 3sinθcosθ = 0 ...(iii) Solving (i) and (iii), we get y = sinθ and x = 2cosθ Now, x2 + 4y2 = 4cos2θ + 4sin2θ = 4 10. (b) : We have, xy = (secφ – tanφ)(cosecφ + cotφ) 1 − sin φ 1 + cos φ = . cos φ sin φ 1 − sin φ + cos φ − sin φ cos φ + sin φ cos φ ⇒ xy + 1 = cos φ sin φ 1 − sin φ + cos φ = ...(i) cos φ sin φ Also, x – y = (secφ – tanφ) – (cosecφ + cotφ) 1 − sin φ 1 + cos φ sin φ − sin2 φ − cos φ − cos2 φ − = cos φ sin φ cos φ sin φ sin φ − cos φ − 1 ...(ii) = cos φ sin φ

∴

18. (c) : sin2α + sin2β + sin2γ 2

tan α

=

2

+

2

tan β 2

+

2

tan γ

1 + tan α 1 + tan β 1 + tan2 γ x y z = + + 1+ x 1+ y 1+ z (Here, x = tan2 α, y = tan2 β, z = tan2 γ) =

(x + y + z) + (xy + yz + zx + 2xyz) + xy + yz + zx + xyz (1 + x)(1 + y)(1 + z) =

1 + x + y + z + xy + yz + zx + xyz =1 (1 + x)(1 + y)(1 + z) (∵ xy + yz + zx + 2xyz = 1)

19. (b) : (cos1° + cos179°) + (cos2° + cos178°) + ... + (cos89° + cos91°) + cos90° + cos180° = –1 20. (a) : Let A + B = θ and A – B = φ. k tan A sin A cos B Then tanA = ktanB or = = 1 tan B cos A sin B Applying componendo and dividendo k + 1 sin A cos B + cos A sin B sin(A + B) sin θ = ⇒ = = k − 1 sin A cos B − cos A sin B sin(A − B) sin φ k +1 ⇒ sin θ = sin φ k −1 3π 21. (c) : Given that π < α < i.e., α is in third 2 quadrant. ⎛π α⎞ Now, (4 sin 4 α + sin2 2α) + 4 cos2 ⎜ − ⎟ ⎝4 2⎠ ⎛π α⎞ = (4 sin 4 α + 4 sin2 α cos2 α) + 2 ⋅ 2 cos2 ⎜ − ⎟ ⎝4 2⎠ ⎡ ⎛π ⎞⎤ = 4 sin2 α(sin2 α + cos2 α) + 2 ⎢1 + cos ⎜ − α ⎟ ⎥ ⎝ ⎠⎦ 2 ⎣ = ± 2sinα + 2 + 2sinα On taking –ve, answer is 2 and on taking +ve, answer is 2 + 4sinα. 3π But π < α < , Hence answer is 2 – 4sinα because 2 sinα is –ve in third quadrant. tan 100° + tan 125° 1 − tan 100° tan 125° tan 100° + tan 125° ∴ tan 225° = 1 − tan 100° tan 125°

22. (d) : tan(100° + 125°) =

tan 100° + tan 125° i.e., 1 = 1 − tan 100° tan 125° i.e., tan100° + tan125° + tan100° tan125° = 1 18

MATHEMATICS TODAY | NOVEMBER‘16

23. (b) : Multiplying both sides by (1 – sinA)(1 – sinB)(1 – sinC), we have, (1 – sin2 A)(1 – sin2B)(1 – sin2C) = (1 – sinA)2 (1 – sinB)2 (1 – sinC)2 ⇒ (1 – sinA)(1 – sinB)(1 – sinC) = ±cosA cosB cosC Similarly, (1 + sinA)(1 + sinB)(1 + sinC) = ±cosA cosB cosC tan α + tan β 1 − tan α tan β 1 1 + x +1 1 1+ x 1+ 2 2 ⇒ tan(α + β) = 1 1 × 1− x 1 + 1 / 2 1 + 2 x +1

24. (b) : tan(α + β) =

⇒ tan(α + β) =

2x + 2 ⋅ 2x + x + 2x + 1

1 + 2x + 2 ⋅ 2x + 2 ⋅ 2x + x − 2x π ⇒ tan(α + β) = 1 ⇒ α + β = 4 25. (a) : S = sinθ + sin2θ + sin3θ + ... + sinnθ We know, sinθ + sin(θ + β) + sin(θ + 2β) + ... n terms nβ sin 2 sin ⎡θ + (n − 1) β ⎤ = ⎢⎣ β 2 ⎥⎦ sin 2 nθ θ(n + 1) sin .sin 2 2 Put β = θ, then S = θ sin 2 26. (b) : Now, cos 70° + 4 sin 70° cos 70° cot 70° + 4 cos 70° = sin 70° cos 70° + 2 sin 140° cos 70° + 2 sin(180° − 40°) = = sin 70° sin 70° sin 20° + sin 40° + sin 40° 2 sin 30° cos 10° + sin 40° = = sin 70° sin 70° sin 80° + sin 40° 2 sin 60° cos 20° = = = 3 sin 70° sin 70° 27. (c) : tan20° tan40° tan60° tan80° =

sin 20° sin 40° sin 80° tan 60° cos 20° cos 40° cos 80°

Let N r = (sin20° sin40° sin80°) =

sin 20° (2 sin 40° sin 80°) 2

=

sin 20° (cos 40° − cos 120°) 2

1 1⎞ ⎛ = sin 20° ⎜1 − 2 sin2 20° + ⎟ ⎝ 2 2⎠ 1 3 ⎛3 ⎞ sin 60° = sin 20° ⎜ − 2 sin2 20° ⎟ = = ⎝2 ⎠ 2 4 8 Now, we take

Dr

=

= cos20° cos40° cos80°

=

sin 2 20°

sin 160° sin 20° 1 = 3 = = = 2 sin 20° 8 sin 20° 8 sin 20° 8

=

30. (c) : tan9° – tan27° – tan63° + tan81° = tan9° – tan27° – cot27° + cot9° = (tan9° + cot9°) – (tan27° + cot27°) 2 2 = − sin 18° sin 54° ⎧⎪ sin 54° − sin 18° ⎪⎫ 2 ⋅ cos 36°⋅ sin 18° =4 = 2⎨ ⎬ = 2. sin118° sin 54° ⎩⎪ sin 18° sin 54° ⎪⎭ sin 3θ + sin 5θ + sin 7θ + sin 9θ 31. (c) : cos 3θ + cos 5θ + cos 7θ + cos 9θ (sin 3θ + sin 9θ) + (sin 5θ + sin 7θ) = (cos 3θ + cos 9θ) + (cos 5θ + cos 7θ) =

2 sin 6θ cos 3θ + 2 sin 6θ cos θ 2 cos 6θ cos 3θ + 2 cos 6θ cos θ

= tan 6θ

sin(B + A) + cos(B − A)

sin(B − A) + cos(B + A) sin(B + A) + sin(90° − B − A)

sin(B − A) + sin(90° − A + B) 2 sin(A + 45°)cos(45° − B)

2 sin(45° − A)cos(45° − B) sin(A + 45°) cos A + sin A = = sin(45° − A) cos A − sin A

3 /8 Hence tan 20° tan 40° tan 80° = 1/ 8

Therefore tan20° tan40° tan60° tan80° = 3 ⋅ 3 = 3 28. (d) : sin36° sin72° sin108° sin144° 1 = sin2 36° sin2 72° = {(2 sin2 36°) (2 sin2 72°)} 4 1 = {(1 − cos 72°) (1 − cos 144°)} 4 1 = {(1 − sin 18°) (1 + cos 36°)} 4 1 ⎡⎛ 5 −1⎞ ⎛ 5 + 1 ⎞ ⎤ 20 1 5 1+ = ⎢ ⎜1 − ⎥= × = ⎟ ⎜ 4 ⎢⎣ ⎝ 4 ⎠⎝ 4 ⎟⎠ ⎥⎦ 16 4 16 29. (b) : x = cos10° cos20° cos40° 1 = [2 sin 10° cos 10° cos 20° cos 40°] 2 sin 10° 1 = [2 sin 20° cos 20° cos 40°] 2 ⋅ 2 sin 10° 1 1 = [2 sin 40° cos 40°] = (sin 80°) 2 ⋅ 4 sin 10° 8 sin 10° 1 1 = cos 10° = cot 10° 8 sin 10° 8

2 cos 6θ (cos 3θ + cos θ)

32. (b) :

3

∴

2 sin 6θ (cos 3θ + cos θ)

33. (a) :

m tan(120° + θ) = n tan(θ − 30°)

m + n tan(θ + 120°) + tan(θ − 30°) = m − n tan(θ + 120°) − tan(θ − 30°) (By componendo and dividendo) sin(θ + 120°)cos(θ − 30°) + cos(θ + 120°)sin(θ − 30°) = sin(θ + 120°)cos(θ − 30°) − cos(θ + 120°)sin(θ − 30°) sin(2θ + 90°) cos 2θ = = = 2 cos 2θ sin(150°) 1/ 2

⇒

34. (b) : tan20° + 2tan50° – tan70° sin 20° sin 70° = − + 2 tan 50° cos 20° cos 70° sin 20° cos 70° − cos 20° sin 70° + 2 tan 50° cos 20° cos 70° sin(20° − 70°) = + 2 tan 50° 1 [cos(70° + 20°) + cos(70° − 20°)] 2 2 sin(−50°) = + 2 tan 50° cos 90° + cos 50° −2 sin 50° = + 2 tan 50° 0 + cos 50° = –2tan50° + 2tan50° = 0 35. (b) : We have, cos(A + C) cos A cos C − sin A sin C cos 2B = = cos(A − C) cos A cos C + sin A sin C 1 − tan2 B 1 − tan A tan C = ⇒ 1 + tan2 B 1 + tan A tan C ⇒ 1 + tan2 B − tan A tan C − tan A tan C tan2 B = 1 − tan2 B + tan A tan C − tan A tan C tan2 B ⇒ 2 tan2 B = 2 tan A tan C ⇒ tan2 B = tan A tan C Hence, tan A, tan B and tan C will be in G.P. =

MATHEMATICS TODAY | NOVEMBER‘16

19

2

tan θ ⎛ sin θ ⎞ = 36. (a, b, c, d) : Since, ⎜ ⎟ ⎝ sin φ ⎠ tan φ sin θ sin θ sin θ cos φ = sin φ sin φ sin φ cos θ sin θ cos φ ⇒ = ⇒ sin 2θ = sin 2φ sin φ cos θ ⇒

⇒ ⇒ 37. (a) ⇒ ⇒ (b)

(c) (d) 38. (a)

2 tan θ 2

=

2 tan φ 2

⇒

6 tan φ 2

=

2 tan φ

1 + tan θ 1 + tan φ 1 + 9 tan φ 1 + tan2 φ 1 tan2 φ = and tan θ = ± 3 3 (a, b, c) : 3α = 2α + α tan3α = tan(2α + α) tan3α – tan2α – tanα = tanα tan2α tan3α sin 4α + sin 2α 2 sin 3α cos α R.H.S = = sin 2α sin 4α sin 2α sin 4α π 3π 2 sin cos 7 7 = 1 = cosecα = 2π 4π sin α sin sin 7 7 π 2π 3π cos α − cos 2α + cos 3α = cos − cos + cos 7 7 7 = 1/2 8cosα cos2α cos4α = –1 (a, b, c, d) : 11π 5π ⎛π⎞ ⎛π⎞ sin sin = sin ⎜ ⎟ cos ⎜ ⎟ ⎝ 12 ⎠ ⎝ 12 ⎠ 12 12 1 ⎛π⎞ 1 = sin ⎜ ⎟ = ∈Q 2 ⎝6⎠ 4

⎛ 9π ⎞ ⎛ 4 π ⎞ ⎛ π ⎞ ⎛π⎞ (b) cosec ⎜ ⎟ sec ⎜ ⎟ = −cosec ⎜⎝ ⎟⎠ sec ⎜ ⎟ = −4 ∈Q ⎝5⎠ ⎝ 10 ⎠ ⎝ 5 ⎠ 10 1 3 2 ⎛π⎞ 2 ⎛π⎞ (c) 1 − 2 sin ⎜ ⎟ cos ⎜ ⎟ = 1 − = ∈Q ⎝8⎠ ⎝8⎠ 4 4 π⎞⎛ 2π ⎞ ⎛ 4π ⎞ 1 ⎛ (d) ⎜ 2 cos2 ⎟ ⎜ 2 cos2 ⎟ ⎜ 2 cos2 ⎟ = ∈Q ⎝ 9⎠⎝ 9 ⎠⎝ 9 ⎠ 8 2 39. (a, b, c) : 2cosec x ( y − 1)2 + 1 ≤ 2

π 3π , and y = 1 2 2 40. (a, c) : Since, α and β satisfy xcosθ + ysinθ = 2a ⇒ (x2 + y2)cos2θ – 4axcosθ + (4a2 – y2) = 0 ⇒ cosec2x = 1 and y = 1 ⇒ x =

4ax 4a2 − y 2 cos α + cos β = 2 , cos α ⋅ cos β = x + y2 x2 + y2 20

MATHEMATICS TODAY | NOVEMBER‘16

⎛α ⎞ ⎛β⎞ ⎛α⎞ ⎛β ⎞ 2 sin ⎜ ⎟ sin ⎜ ⎟ = 1 ⇒ 4 sin2 ⎜ ⎟ sin2 ⎜ ⎟ = 1 ⎝ 2 ⎠ ⎝2⎠ ⎝2⎠ ⎝2⎠ ⇒ cosα + cosβ = cosα ⋅ cosβ 4ΔR abc 41. (a, b, c, d) : R ≥ 2r ⇒ R2 ≥ ⇒ R2 ≥ 2S a +b+c If ∠C = 90° ⇒ a2 + b2 = c2 , c = 2R C r + 2R = (s – c) tan + c = s – c + c = s 2 π 3π 5π 42. (b) : sec ,sec ,sec 7 7 7 are the roots of x3 – 4x2 – 4x + 8 = 0 ⎛π⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ sec ⎜ ⎟ + sec ⎜ ⎟ + sec ⎜ ⎟ = 4 ⎝7⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ 43. (b) : 8x 3 − 4x 2 − 4x + 1

π⎞⎛ 3π ⎞ ⎛ 5π ⎞ ⎛ = 8 ⎜ x − cos ⎟ ⎜ x − cos ⎟ ⎜ x − cos ⎟ ...(i) ⎝ 7 ⎠⎝ 7 ⎠⎝ 7 ⎠ ⎛ π ⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ 1 Put x = 1 ⇒ sin ⎜ ⎟ sin ⎜ ⎟ sin ⎜ ⎟ = ⎝ 14 ⎠ ⎝ 14 ⎠ ⎝ 14 ⎠ 8 44. (d) : Put x = –1 in (i), we get 7 ⎛π⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ ⇒ cos ⎜ ⎟ cos ⎜ ⎟ cos ⎜ ⎟ = ⎝ 14 ⎠ ⎝ 14 ⎠ ⎝ 14 ⎠ 8 (45 - 47) : 45. (a) 46. (b) sinα + sinβ = 1/ 4 ⎛α −β⎞ ⎛α +β⎞ cos ⎜ = 1/ 4 ⇒ 2 sin ⎜ ⎟ ⎝ 2 ⎟⎠ ⎝ 2 ⎠ cosα + cosβ = 1/3 ⎛α −β⎞ ⎛α +β⎞ = 1/ 3 cos ⎜ ⇒ 2 cos ⎜ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ Dividing (iv) by (ii), we have ⎛α +β⎞ 3 = tan ⎜ ⎝ 2 ⎟⎠ 4

47. (d) ...(i) ...(ii) ...(iii) ...(iv)

⎛α +β⎞ 3 2 tan ⎜ 2× ⎝ 2 ⎟⎠ 4 = 24 = ⇒ sin(α + β) = 2 25 ⎛α +β⎞ ⎛3⎞ 1 + tan2 ⎜ ⎟ 1 + ⎜⎝ ⎟⎠ ⎝ 2 ⎠ 4 2 ⎛3⎞ ⎛α +β⎞ 1 − tan2 ⎜ 1− ⎜ ⎟ ⎟ ⎝4⎠ ⎝ 2 ⎠ 7 = = And cos(α + β) = 2 ⎛α +β⎞ 25 ⎛3⎞ 1 + tan2 ⎜ ⎝ 2 ⎟⎠ 1 + ⎜⎝ 4 ⎟⎠ 48. A → R; B → S; C → P; D → Q 1 (A) A = 20° ⇒ 3A = 60° ⇒ cos 3A = 2 ⇒ 8x3 – 6x – 1 = 0 where x = cos20°

(B) ⇒ (C) ⇒ (D) ⇒

A = 10° ⇒ sin3A = 1/2 8x3 – 6x + 1 = 0 where x = sin10° A = 15° ⇒ tan3A = 45° x3 – 3x2 – 3x + 1 = 0 where x = tan15° A = 6° ⇒ sin5A = 1/2 32x5 – 40x3 + 10x – 1 = 0 where x = sin6°

49. A → S; B → R; C → Q; D → P (A) cos(2A + θ) + cos(2B + θ) = 2cos(A + B + θ) cos(A – B) ≤ 2cos(A – B) (B) cos2A + cos2B = 2cos(A + B) cos(A – B) ≤ 2cos(A + B) (C) y = secx always concave up sec 2A + sec 2B ∴ ≥ sec(A + B) 2 (D) tan θ + cot θ − 2 cos(2A + 2B) = ( tan θ + cot θ)2 + 4 sin2(A + B) ≥ 2sin(A + B) 50. (A) ⇒ ⇒ ⇒ ⇒ ⇒ (B) ∴ ⇒ ∴

A → Q; B → S; C → P We have, Δ = a2 – (b – c)2 Δ = a2 – b2 – c2 + 2bc b2 + c2 – a2 = 2bc – Δ 1 2bc cos A = 2bc − bc sin A ⇒ 4cosA + sinA = 4 2 A A ⎛ 2 A⎞ 4 ⎜1 − 2 sin ⎟⎠ + 2 sin cos = 4 ⎝ 2 2 2 A 1 8 tan = ⇒ tan A = 2 4 15 Let tan A = 3k, tanB = 4k, tanC = 5k tanA + tanB + tanC = tanA tanB tanC 1 12k = 60 k3 ⇒ k = 5 3 4 tan A = , tan B = , tan C = 5 5 5

∴ sin A sin B sin C = (C) We have, f (θ) =

2 5 7 2 cos2 θ − 2 sin θ cos θ 2

2

52. (1) : 4x2 – 16x + 15 < 0 4x2 – 10x – 6x + 15 < 0 2x(2x – 5) – 3(2x – 5) < 0 3 5 ⇒ 0 b L(x1, y1) 0 or L(x1, y1) . L(x2, y2) > 0 L(x2 , y2 ) Then the point P(x1, y1) and Q(x2, y2) lie on same side of line ax + by + c = 0. (ii) If ax1 + by1 + c and ax2 + by2 + c both are of opposite L(x1, y1) in sign or 0 (b) The obtuse angle, if c1c2 ANGLE BISECTOR Consider two lines L1 ≡ a1x + b1y + c1 = 0 and L2 ≡ a2x + b2 y + c2 = 0 Let P(x, y) be any point on either of bisectors. ⊥r Distance of P from L1 = ⊥ r Distance of P from L2   

  






          

         

∴

| a1x + b1 y + c1 | | a2 x + b2 y + c2 | = a12 + b12 a22 + b22

(

)

(

)

PROBLEMS Single Correct Answer Type

1. ABC is a variable triangle with the fixed vertex C(1, 2) and A, B having the coordinates (cost, sint), (sint, –cost) respectively, where t is a parameter. The locus of the centroid of the ΔABC is (a) 3(x2 + y2) – 2x – 4y – 1 = 0 (b) 3(x2 + y2) – 2x – 4y + 1 = 0 (c) 3(x2 + y2) + 2x + 4y – 1 = 0 (d) 3(x2 + y2) + 2x + 4y + 1 = 0 2. The lines x + 2y + 3 = 0, x + 2y – 7 = 0 and 2x – y – 4 = 0 are the sides of a square. Equation of the remaining side of the square can be (a) 2x – y – 14 = 0 (b) 2x – y + 8 = 0 (c) 2x – y – 10 = 0 (d) 2x – y – 6 = 0 3. If in a ΔABC (whose circumcentre is at the origin), and a ≤ sinA, then for any point (x, y) inside the circumcircle of ΔABC

(a) | xy | < (c)

1 8

1 1 < xy < 8 2

(b) | xy | >

1 8

(d) none of these

4. If the vertices of a triangle are A(10, 4), B(–4, 9) and C(–2, –1), then the equation of its altitudes are (a) x – 5y + 10 = 0, 12x + 5y + 3 = 0 and 14x + 5y – 23 = 0 (b) 14x – 5y + 10 = 0, x + 5y + 3 = 0 and 14x – 5y + 23 = 0 (c) x – 5y + 1 = 0, 12x + 5y + 3 = 0 and x – 5y + 23 = 0 (d) x – 5y + 10 = 0, 12x + 5y + 3 = 0 and 14x – 5y + 23 = 0 5. If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x secθ + y cosecθ = a and x cosθ – y sinθ = a cos2θ respectively, then the value of 4p12 + p22 is (a) 4a2

(b) 2a2

(c) a2

(d) none of these

6. If (1, 1) and (–3, 5) are vertices of a diagonal of a square, then the equations of its sides through (1, 1) are (a) 2x – y = 1, y – 1 = 0 (b) 3x + y = 4, x – 1 = 0 (c) x = 1, y = 1 (d) none of these 7. A family of lines is given by (1 + 2λ)x + (1 – λ)y + λ = 0, λ being the parameter. The line belonging to this family at the maximum distance from the point (1, 4) is (a) 4x – y + 1 = 0 (b) 33x + 12y + 7 = 0 (c) 12x + 33y = 7

(d) none of these

8. If A(sin α, 1 / 2 ) and B(1 / 2 , cos α) , –π ≤ α ≤ π, are two points on the same side of the line x – y = 0 then α belongs to the interval ⎛ π π ⎞ ⎛ π 3π ⎞ ⎛ π π⎞ (a) ⎜ − , ⎟ ∪ ⎜ , ⎟ (b) ⎜ − , ⎟ ⎝ 4 4⎠ ⎝4 4 ⎠ ⎝ 4 4⎠ π 3 π ⎛ ⎞ (c) ⎜ , (d) none of these ⎝ 4 4 ⎟⎠ 9. If the point (cosθ, sinθ) does not fall in that angle between the lines y = |x – 1| in which the origin lies, then θ belongs to ⎛ π 3π ⎞ (a) ⎜ , ⎝ 2 2 ⎟⎠ (c) (0, π)

⎛ π π⎞ (b) ⎜ − , ⎟ ⎝ 2 2⎠ (d) none of these

10. If a ray travelling along the line x = 1 gets reflected from the line x + y = 1, then the equation of line along which the reflected ray travel is (a) y = 0 (b) x – y = 1 (c) x = 0 (d) none of these 11. The point P(2, 1) is shifted by 3 2 parallel to the line x + y = 1, in the direction of increasing ordinate, to reach Q. The image of Q by the line x + y = 1 is (a) (5, –2) (b) (–1, –2) (c) (5, 4) (d) (–1, 4) 12. In triangle ABC; A(1, 1), B(4, –2), C(5, 5). Equation of the internal angle bisector of ∠A is (a) y = 1 (b) x – y = 1 (c) y = 5 (d) x + y = 5 Multiple Correct Answer Type

13. A line which makes an acute angle θ with the positive direction of x-axis is drawn through the point P(3, 4) to meet the line x = 6 at R and y = 8 at S, then (a) PR = 3 secθ (b) PS = 4 cosecθ 2(3 sin θ + 4 cos θ) (c) PR + PS = sin 2θ 9 16 + =1 (d) (PR)2 (PS)2 14. If the lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent (a + b + c ≠ 0), then (a) a3 + b3 + c3 – 3abc = 0 (b) a = – b (c) a = b = c (d) a2 + b2 + c2 – bc – ca – ab = 0 15. If 6a2 – 3b2 – c2 + 7ab – ac + 4bc = 0, then the family of lines ax + by + c = 0 is concurrent at (a) (–2, –3) (b) (3, –1) (c) (2, 3) (d) (–3, 1) 16. If (α, α2) lies inside the triangle formed by the lines 2x + 3y – 1 = 0, x + 2y – 3 = 0, 5x – 6y – 1 = 0, then (a) 2α + 3α2 – 1 > 0 (b) α + 3α2 – 3 < 0 (c) α + 2α2 – 3 < 0 (d) 6α2 – 5α + 1 > 0 17. A(1, 2) and B(7, 10) are two points. If P(x, y) is a point such that the angle APB is 60° and the area of the triangle APB is maximum, then which of the following is/are true? (a) P lies on any line perpendicular to AB (b) P lies on the right bisector of AB (c) P lies on the straight line 3x + 4y = 36 (d) P lies on the circle passing through the points (1, 2) and (7, 10) and having a radius of 10 units MATHEMATICS TODAY | NOVEMBER‘16

25

⎛ a 3 a 2 − 3 ⎞ ⎛ b3 b2 − 3 ⎞ 18. If the points ⎜ , , ⎟ and ⎟, ⎜ ⎝ a −1 a −1 ⎠ ⎝ b −1 b −1 ⎠ ⎛ c3 c2 − 3 ⎞ , ⎟ , where a ≠ b ≠ c ≠ 1, lies on the line ⎜ ⎝ c −1 c −1 ⎠ lx + my + n = 0 , then (a) a + b + c = –m/l (b) ab + bc + ca = n/l m+n (c) abc = l (d) abc – (bc + ca + ab) + 3(a + b + c) = 0 Comprehension Type

Paragraph for Q. No. 19 to 21 Let A(0, β), B(–2, 0) and C(1, 1) be the vertices of a triangle then 19. Angle A of the triangle ABC will be obtuse if β lies in ⎛ 5⎞ (a) (–1, 2) (b) ⎜ 2, ⎟ ⎝ 2⎠ 2 2 (c) ⎜⎛ −1, ⎟⎞ ∪ ⎛⎜ , 2 ⎟⎞ ⎝ 3⎠ ⎝3 ⎠

(d) none of these

20. If I1 is the interval of values of β for which A is obtuse and I2 be the interval of values of β for which A is largest angle of ΔABC, then (a) I1 = I2 (b) I1 is a subset of I2 (c) I2 is a subset of I1 (d) none of these

(a) 61 (c) 100

(b) 22 (d) 36 Paragraph for Q. No. 25 to 27

The vertex A of triangle ABC is (3, –1). The equations of median BE and angular bisector CF are 6x + 10y – 59 = 0 and x – 4y + 10 = 0. Then 25. Slope of the side BC must be (a) 1/9 (b) –2/9 (c) 1/7 (d) none of these 26. The equation of AB must be (a) x + y = 2 (b) x + 4y = 0 (c) 18x + 13y = 41 (d) 23x – y = 70 27. The length of the side AC must be (a)

71

(b)

(c)

85

(d) none of these

83

Integer Answer Type

28. The number of integral values of m for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is 29. The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then the value

21. All the value of β for which angle A of the triangle ABC is largest lie in interval 2⎞ ⎛2 ⎞ ⎛ (a) (–2, 1) (b) ⎜ −2, ⎟ ∪ ⎜ , 1⎟ ⎝ 3⎠ ⎝3 ⎠

1 + 2 of a b is 1 1 + 2 2 p q

2 2 (c) ⎜⎛ −2, ⎟⎞ ∪ ⎛⎜ , 6 ⎞⎟ (d) none of these ⎝ ⎠ 3⎠ ⎝3

30. If λ : 1 is the ratio in which the line joining the points (2, 3) and (4, 1) divides the segment joining the points (1, 2) and (4, 3), then the value of λ is

Paragraph for Q. No. 22 to 24 Given the equations of two sides of a square as 5x + 12y – 10 = 0, 5x + 12y + 29 = 0. Also given a point M(–3, 5) lying on one of its sides. 22. The number of possible squares must be (a) one (b) two (c) four (d) none of these 23. The area of the square must be (a) 9 sq. units (b) 6 sq. units (c) 5 sq. units (d) none of these 24. If the possible equations of the remaining sides is 12x – 5y + λ = 0, then λ cannot be 26

MATHEMATICS TODAY | NOVEMBER‘16

1

2

31. If the quadrilateral formed by the lines ax + by + c = 0, a′x + b′y + c = 0, ax + by + c′ = 0, a′x + b′y + c′ = 0 have perpendicular diagonals, then the value of

a 2 + b2 a ′ 2 + b′ 2

is

32. The lines x + y = | a | and ax – y = 1 intersect each other in the first quadrant. Then the minimum sum of two different integral value of a is 33. A ray of light along x + 3 y = 3 gets reflected upon reaching x-axis and the equation of the reflected ray is ax + by + c = 0 then the value of bc is

Matrix-Match Type

34. Match the following: (A)

(B) (C)

(D)

Column-1 If P(1, 1), Q(4, 2) and R(x, 0) be three points such that PR + RQ is minimum, then x is equal to The area bound by the curves max {|x|, |y|} = 1 is equal to The number of circles that touch all the three lines 2x – y = 5, x + y = 3 and 4x – 2y = 7 is equal to If the point (a, a) lies between the lines |x + y| = 6 then [|a|] can be equal to, ([⋅] represents the integral part)

Column-II (P) 0

(Q)

1

(R)

2

(S)

3

(T)

4

3. (a) : Since, a ≤ sinA ⇒

35. Match the following: Column-1 (A) If the point (x1 + t(x2 – x1), y1 + t(y2 – y1)) divides the join of (x1, y1) and (x2, y2) internally, then (B) Set of values of ‘t’ for which point P(t, t 2 – 2) lies inside the triangle formed by the lines x + y = 1, y = x + 1 and y = –1 is (C) IfP (1 + (t / 2 ), 2 + (t / 2 )) be any point on a line, then the value of t for which the point P lies between parallel lines x + 2y = 1 and 2x + 4y = 15, is (D) If the point (1, t) always remains in the interior of the triangle formed by the lines y = x, y = 0 and x + y = 4, then

Column-II (P)

(Q)

−

4 2 5 2

(ii) ⇒ sin α
0

and cosα >

9. (b) : The lines are y = x – 1 and y = –x + 1 and (cosθ, sinθ) is any point on the circle x2 + y2 = 1, where centre = (0, 0) and radius = 1. Clearly from the figure, θ can vary from π π − to . 2 2 

10. (a) : 

....(i)

− cos α < 0

and cosα
AB2 + AC2 and A, B, C are non-collinear. ⇒ 10 > β2 + 4 + 1 + (β – 1)2 ⇒ (β – 2)(β + 1) < 0 ⇒ β ∈ (–1, 2) But A, B, C are collinear for β = 2/3 2⎞ ⎛2 ⎞ ⎛ ⇒ Correct interval for β is ⎜ −1, ⎟ ∪ ⎜ , 2 ⎟ ⎝ 3⎠ ⎝3 ⎠ 20. (b) : Whenever A is obtuse, it is the largest angle also. But there may be more values of β for which A is largest but is not obtuse. (i.e., right angle at A) ⇒ Choice (b) is correct. 21. (c) : Angle A will be largest if a > b, a > c and A, B, C are non-collinear ⇒ a2 > b2 , a2 > c2 , β ≠ 2/3 ⇒ 10 > 1 + (β – 1)2 , 10 > 4 + β2 ⇒ β2 – 2β – 8 < 0, β2 < 6 ⇒ –2 < β < 4, − 6 < β < 6 On taking intersection and rejecting β = 2/3, we get 2⎞ ⎛2 ⎞ ⎛ β ∈ ⎜ −2, ⎟ ∪ ⎜ , 6 ⎟ ⎠ ⎝ 3⎠ ⎝3

(22-24) :

22. (b) : It is evident that M(–3, 5) lies on some other side and possible squares are obviously two. 23. (a) : The perpendicular distance between AB and CD 29 − (−10) 39 is = = 3 ⇒ Area = (3)2 = 9 sq. units 2 2 13 a +b 24. (d) : The equations of remaining sides can easily be determined as 12x – 5y + 61 = 0, 12x – 5y + 22 = 0, 12x – 5y + 100 = 0 ⇒ (d) is correct. Equation of BC is 12x – 5y + 61 = 0 ⇒ Other possible sides are 12x – 5y + c = 0, where | c − 61 | = 3 ⇒ c = 100, c = 22 13 (25 - 27) :

(3, –1)





 =0 59 – 0  – +1 4 +1 6 0= 0

(, )

25. (b) : Let coordinates of C be (h, k), then h – 4k + 10 = 0 ....(i) h + 3 k −1⎞ and E ⎜⎛ , ⎟ lies on 6x + 10y – 59 = 0 ⎝ 2 2 ⎠ ⎛ k −1⎞ ⎛h+3⎞ − 59 = 0 + 10 ⎜ ⇒ 6⎜ ⎟ ⎝ 2 ⎟⎠ ⎝ 2 ⎠ ⇒ 3h + 5k = 55 From (i) and (ii), we get h = 10, k = 5 ⇒ Coordinates of C are (10, 5) 6 ⇒ Slope of AC = 7 ⇒ Slope of BC must be given by 6 1 1 − m− 6 −2 4 = 7 4 ⇒ m= , 6 1 m 7 9 1+ × 1+ 7 4 4 MATHEMATICS TODAY | NOVEMBER‘16

....(ii)

29

2 Since 6/7 represents slope, we have slope BC = − 9 2 26. (c) : Equation of BC is y − 5 = − (x − 10) 9 or 2x + 9y – 65 = 0 Now the point B can be found by solving BC and BE whence equation of AB can be determined as 18x + 13y = 41 27. (c) 28. (2) : 3x + 4(mx + 1) = 9 5 ⇒ (3 + 4m)x = 5 ⇒ x = 3 + 4m For x to be an integer, 3 + 4m should be a divisor of 5 i.e., 1, –1, 5 or –5. 3 + 4m = 1 ⇒ m = – 1/2 (not an integer) 3 + 4m = –1 ⇒ m = –1 (integer) 3 + 4m = 5 ⇒ m = 1/2 (not an integer) 3 + 4m = –5 ⇒ m = –2 (integer) Hence, there are two integral values of m. 29. (1) : The equation of the line L in the two coordinate x y X Y systems are + = 1 and + = 1 where (X, Y) are a b p q the new coordinates of a point (x, y) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed, 1 1 1 1 1 1 = ⇒ 2+ 2= 2+ 2 1 1 1 1 a b p q + 2 + 2 2 2 a b p q 1 1 + 2 2 ⇒ a b =1 1 1 + 2 2 p q 30. (1) : The equation of the line joining the points (2, 3) and (4, 1) is 1− 3 (x − 2) ⇒ x + y – 5 = 0 ….(i) y −3= 4−2 Suppose the line joining (2, 3) and (4, 1) divides the segment joining (1, 2) and (4, 3) at the point P in the ratio of λ : 1. 4λ + 1 3λ + 2 ⎞ Then the coordinates of P are ⎛⎜ , ⎝ λ + 1 λ + 1 ⎟⎠ Clearly, P lies on (i), 4λ + 1 3λ + 2 ⇒ + −5 = 0 ⇒ λ =1 λ +1 λ +1 31. (1) : Since the diagonals are perpendicular, so the given quadrilateral is a rhombus. ∴ Distance between two pairs of parallel sides are equal. 30

MATHEMATICS TODAY | NOVEMBER‘16

⇒

c′ − c 2

a +b

2

=

c′ − c a′2 + b′2

⇒ a2 + b2 = a′ 2 + b′ 2

32. (3) : Given lines intersect at the point ⎛ 1+ | a | a | a | −1⎞ P⎜ , ⎝ 1 + a 1 + a ⎟⎠ This point will lie in first quadrant iff 1 + a > 0 and a | a | – 1 ≥ 0 ⇒ a > – 1 and a | a | – 1 ≥ 0 ⇒ a2 – 1 ≥ 0 if a ≥ 0 or – a2 – 1 > 0 if 0 < a < – 1 ⇒ a ∈ [1, ∞) 33. (3) : The line x + 3 y = 3 cuts y-axis at P(0, 1). Clearly its image Q(0, –1) lies on the reflected ray AR produced backward.

So, the equation of the reflected ray AR is 0 +1 (x – 0) or 3 y = x − 3 y+1= 3 −0 34. (A) – (R); (B) – (T) ; (C) – (R) ; (D) – (P,Q,R)  (A) The image of P in the x-axis   is P′(1, –1). Equation of P ′Q    3  is y + 1 = (x – 1), which  3  intersects x–axis at (2, 0). (B) The region is a square of side length 2, so the desired area is 4.

(C) As two of the lines are parallel, so two such circles are possible. (D) The line x = y cuts the lines x + y = ± 6 at (–3, –3) and (3, 3) ⇒ – 3 < a < 3. ∴ [| a |] = 0, 1, 2. 35. (A) – (R) ; (B) – (Q) ; (C) – (P) ; (D) – (R) ““

CLASS XI

Series 7

 Straight Lines | Conic Sections | Introduction to Three Dimensional Geometry STRAIGHT LINES DEFINITION A path traced by a point in a constant direction and endlessly in its opposite direction is called straight line. Distance between A(x1, y1) and B(x2, y2) is

Distance Formula Internal Division Section Formula

External Division

Mid Point Formula

Area of a Triangle

AB = (x2 − x1 )2 + ( y2 − y1 )2 Coordinates of M, if M divides the join of A(x1, y1) and B(x2, y2) internally mx + nx1 my2 + ny1 ⎞ in the ratio m : n is ⎛⎜ 2 , ⎟ ⎝ m+n m+n ⎠ Coordinates of M, if M divides the join the of A(x1, y1) and B(x2, y2) ⎛ mx − nx1 my2 − ny1 ⎞ externally in the ratio m : n is ⎜ 2 , ⎟ ⎝ m−n m−n ⎠ Coordinates of M, if M bisects the join of A(x1, y1) and B(x2, y2) is ⎛ x1 + x2 y1 + y2 ⎞ , ⎟ ⎜⎝ 2 2 ⎠ Area of ΔABC with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is 1 Δ = x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) 2

SLOPE OF A LINE If a line makes an angle θ with the positive direction of x-axis, then tan θ is called the slope or gradient of the line. It is denoted by m. Slope of a line Value of m Parallel to x-axis

0

Parallel to y-axis

not defined

Passing through two points (y2 – y1)/(x2 – x1) (x1, y1) and (x2, y2)

CONDITION OF PARALLELISM AND PERPENDICULARITY Two lines having slopes m1 and m2 are said to be (i) Parallel iff m1 = m2 (ii) Perpendicular iff m1m2 = –1 ANGLE BETWEEN TWO LINES Angle between two lines having slopes m1 and m2 is, m − m2 tan θ = 1 , 1 + m1m2 ≠ 0 1 + m1m2 MATHEMATICS TODAY | NOVEMBER‘16

31

COLLINEARITY OF THREE POINTS Three points A, B, C are said to be collinear, if slope of AB = slope of BC DIFFERENT FORMS OF EQUATION OF LINE Equation Equation of x-axis

y=0

Equation of y-axis

x=0

Equation of line parallel to x-axis

y=a

Equation of line parallel to y-axis

x=b

Point slope form : Equation of line whose slope is m and passing through the point (x1, y1)

y – y1 = m (x – x1)

Figure 







 









     



 

Two point form : Equation of line passing through A(x1, y1) and B(x2, y2)

y −y y – y1 = 2 1 (x – x1) x2 − x1



 

                



Slope intercept form : Equation of line having slope m and cuts an intercept c on y-axis.

y = mx + c

   
    

Intercept form : Equation of line cuts intercepts a and b on x and y-axes respectively.

x y + =1 a b



     
 

Normal form : Equation of line having normal distance from origin p and this normal makes an angle α with the +ve x-axis.

xcosα + ysinα = p





Note : If a line with slope m makes x-intercept d. Then equation of the line is y = m (x – d). GENERAL EQUATION OF A LINE An equation of the form Ax + By + C = 0, where A, B, C are constants and A, B are not simultaneously zero is called the general equation of a line. 32

MATHEMATICS TODAY | NOVEMBER‘16



 

 

Different forms of Ax + By + C = 0 A C Slope - intercept form : y = − x − , B ≠ 0  B B y x + = 1, C ≠ 0 Intercept form :  −C / A −C / B Normal form : x cos α + y sin α = p 



where cos α = ± p=±

A A2 + B 2

, sin α = ±

B A2 + B 2

C 2

A + B2

Distance of a point from a line : Distance of a point (x1, y1) from a line Ax + By + C = 0 is, d=

Ax1 + By1 + C

A2 + B 2 Distance between two parallel lines : Distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 C − C1 is, d = 2 A2 + B 2 E Q UAT I O N O F FA M I LY O F L I N E S PA S S I N G THROUGH THE POINT OF INTERSECTION OF TWO LINES Let the two intersecting lines be A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 So, equation of line passing through intersection of above lines is A1x + B1y + C1 + λ(A2x + B2y + C2) = 0 where λ is an arbitrary constant called parameter.

SHIFTING OF ORIGIN Translation of axes An equation corresponding    
  to a set of points with reference to a system of coordinate axes may be    simplified by taking the  s e t o f p o i nt s i n s o m e  other suitable coordinate 

 system such that all geometric properties remain unchanged. We can form a transformation in which the new axes can be transformed parallel to the original axes and origin can be shifted to a new point. This kind of transformation is called a translation of axes. The coordinates of each point of the plane are changed under a translation of axes. To see how coordinates of a point of the plane changed under a translation of axes, let us take a point P(x, y) referred to the axes OX and OY. Let O′X′ and O′Y′ be new axes parallel to OX and OY respectively, where O′ is the new origin. Let (h, k) be the coordinates of O′ referred to the old axes, i.e., OL = h and LO′ = k. Also, OM = x and MP = y The transformation relation between new coordinates (x′, y′) and old coordinates (x, y) are given by x = x′ + h, y = y′ + k

CONIC SECTIONS

   

DEFINITION The locus of a point which moves in a plane such that the ratio of its distance from a fixed point to its perpendicular distance from a fixed straight line is always constant, is known as a conic section or a conic. The fixed point is called   fo cus of the conic, f ixed line is called directrix of   the conic and the constant ratio is called eccentricity of the conic. CIRCLE, PARABOLA, ELLIPSE AND HYPERBOLA When the plane cuts the nappe (other than the vertex) of the cone, we have the following situations. Relationship/ Conic α : S em i ver t i c a l Value of angles angle of right circular cone. Circle β = 90° β : Angle made by Ellipse α < β < 90° the intersecting plane Parabola β=α with the vertical axis Hyperbola of the cone. 0