(1026)Dpp 6 Solutions of Triangle and Conic Section b.pdf.Tmp

8/17/2019 (1026)Dpp 6 Solutions of Triangle and Conic Section b.pdf.Tmp

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DATE : 24.04.2016 CUMULATIVE TEST-03 (CT-03)Syllabus : Inverse Trigonometric Function & Limit Continuity & Derivability, Quadratic Equation, Applicationof Derivatives, Sequence and Series, Binomial Theorem, Straight line, Circle

TARGET : JEE (ADVANCED) 2016O

Course : VIJETA (ADP) & VIJAY (ADR) Date : 24-04-2016

MM AA TT HH EE MM AA TT II CC SS

DPPDPPDPPDAILY PRACTICE PROBLEMS

NO. 06

TT EE SS TT II NN FF OO RR MM AA TT II OO NN

REVISION DPP OFS OLUTION OF TRIANGLE AND CONIC S ECTION

Total Marks : 147 Max. Time : 116 min.Single choice Objective ('–1' negative marking) Q.1 to Q.12 (3 marks 3 min.) [36, 36]Multiple choice objective ('–1' negative marking) Q.13 to Q.28 (4 marks 3 min.) [64, 48]

Subjective Questions ('–1' negative marking) Q.29 to Q.33 (3 marks 3 min.) [15, 15]Comprehension ('–1' negative marking) Q.34 to Q.40 (3 marks 3 min.) [21, 21]

1. In a triangle ABC, if 2015c 2 = a 2 + b 2 and cot C = N(cot A + cot B), then the number of distinct primefactor of N is ABC esa ;fn2015c 2 = a 2 + b 2 rFkkcot C = N(cot A + cot B) rc N ds fofHkUu vHkkT; Hkktdks a dh la [;k gS&

(A) 0 (B) 1 (C*) 2 (D) 4

Sol. cotC = N(cotA + cotB) cosC cos A cosB

NsinC sin A sinB

2 2 2 2 2 2 2 2 2a b – c b c – a a c – b

N4 4 4

N = 1007 = 19 × 53

2. The number of right angle t`riangles of integer side lengths whose product of leg lengths is equal tothree times the perimeter is

iw.kkZa d Hkqtkvksa dh yEckbZ okys ,sls ledks .k f=kHkqtks dh la[;k ftues a ledks .k 'kh"kZ cukus okyh nks uks a Hkqtkvksa dk xq.kuQy ifjeki dk3 xquk gS] gksxh&(A) 0 (B) 1 (C) 2 (D*) 3

Sol.

3

3

a–3

a–3

b–3b–3

ANSWERKEY OF DPP # 061. (C) 2. (D) 3. (A) 4. (C) 5. (A) 6. (A)7. (C) 8. (B) 9. (A) 10. (C) 11. (D) 12. (A)

13*. (CD) 14. (CD) 15. (ABD) 16. (AB) 17. (AC) 18. (BC)19. (AD) 20. (AC) 21. (BC) 22. (ABD) 23. (ACD) 24. (AC)25. (AC) 26. (ABC) 27. (ABD) 28. (ABC) 29. 8 30. 831. 4 32. 5 33. 8 34. (B) 35. (A) 36. (C)37. (B) 38. (D) 39. (ABD) 40. (ABCD)


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ab = 6s 2 = 6s r = 3Now vc , a 2 + b 2 = (a + b – 6) 2 (a – 6) (b – 6) = 18

3_. An ellipse and hyperbola share common foci F 1, and F 2. The ellipse has vertices at the origin and (0,24) and a semi-minor axis of length 11. The hyperbola has a conjugate axis of length 4 3 . The ellipseand hyperbola intersect at four points. Let P be one of the points of intersection, then value of (PF 1)

2 +(PF 2)

2 is. ,d nh?kZ o`Ùk rFkk vfrijoy; dh ukfHk;kaF1, rFkkF2 mHkfu"B gsA nh?kZo`Ùk ds 'kh"kZ ewy fcUnq rFkk(0, 24) ij gS vkS jv)Zy?kq v{k dh yEckbZ11 gS] rFkk vfrijoy; la ;qXeh v{k dh yECkkbZ4 3 gSA nh?kZo`Ùk vkSj vfrijoy; pkj

fcUnqvka s ij izfrPNs n djrs gSA ekuk izfrPNsn fcUnqvks a es a ls ,d fcUnqP gSA rc(PF 1)2 + (PF 2)2 dk eku gSA

(A*) 410 (B) 820 (C) 532 (D) 266Sol. For ellipse PF 1 + PF 2 = major axis

PF 1 + PF 2 = 26 (i.e. distance between vertices)= 2a a = 13

for an ellipse distance between centre and focus = ae = 2 2a b = 169 121 = 4 3ellipse and hyperbola are con-focal.

distance between centre and focus for hyperbola = 4 3

4 3 = Ae'. 4 3 =2 2

A B A2

=2 2

(4 3 ) (2 3) A = 6for hyperbola, |PF 1 – PF 2| = 2A = 12

PF 1 + PF 2 = 26 & |PF 1 – PF 2 | = 12square and add, we get.(PF 1)

2 + (PF 2)2 = 410.

nh?kZ o`Ùk ds fy, PF 1 + PF 2 = nh?kZv{kPF 1 + PF 2 = 26 ( 'kh"kksZ ds e/; nwjh)

= 2a a = 13

nh?kZ Zo`Ùk ds fy, ukfHk vkS j dsUnz ds e/; nwjh= ae = 2 2a b = 169 121 = 4 3 nh?kZ o`Ùk vkS j vfrijoy; dh ukfHk;ka laikrh gS

vfrijoy; ds fy, ukfHk vkS j dsUnz ds e/; nwjh= 4 3 4 3 = Ae'. 4 3 = 2 2 A B A2 = 2 2(4 3 ) (2 3)

A = 6vfrijoy; ds fy, , |PF 1 – PF 2 | = 2A = 12

PF 1 + PF 2 = 26 & |PF 1 – PF 2 | = 12 oxZ djds tks Mus ij(PF 1)

2 + (PF 2)2 = 410.

4_. If 'O' is the circumcentre of ABC and R 1, R 2 & R3 are the radii of the circumcircles of triangles OBC,

OCA and OAB respectively. then1 2 3

a b c

R R R

has the value.

;fn 'O', ABC dk ifjdsUnz gS rFkk f=kHkq tks aOBC, OCA vkS j OAB ds ifjxr o`Ùkks a dh f=kT;ka,a Øe'k%R1, R 2 & R 3

gS] rc1 2 3

a b cR R R

dk eku gSA

(A)3

abc

2R (B)

3Rabc

(C*)2

4

R (D)

24R


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(A*) ijoy; ds v{k ds lekUrj(B) ijoy; ds v{k ds yEcor~ (C) ijoy; ds v{k ds uk rks lekUrj vkS j uk gh yEcor (D) dqN ugh dgk tk ldrk

Hint. orthocentre lies on directrix yEcdsUæ fu;rk ij fLFkr gS

H(–a, a(t – 1/t)) & 2 2a 1 1

G t – 1 ,a t –3 tt

centroid dsUæd

7. Normals AO, AA 1 & AA2 one drawn to the parabola y2 = 8x from A(h, 0). If triangle OA 1 A2 is equilateral,

then ‘h’ can be equal to fcUnq A(h, 0) ls ijoy; y2 = 8x ij vfHkyEc AO, AA 1 rFkk AA2 [khaps x;s gSA ;fn f=kHkqtOA 1 A2 ,d leckgq f=kHkqt gS] rc‘h’ dk eku gks ldrk gSµ (A) 24 (B) 26 (C*) 28 (D) 30

Hint. 0

A1(t1)

30°

A2(–t 1)

A(h, 0)

equation of normal vfHkyEc dk lehdj.k y = – tx + 2at + at 3 ...(i)

&1

2t

= tan1

6 3 t1 = 2 3 ...(ii)

equation lehdj.k (ii) & vkSj (i) h = 28

8. The auxilliary circle of a family of ellipse passes through origin and makes intercepts of 8 and 6 units on

the x-axis and y-axis respectively. If the eccentricity of all such ellipses is12

, then the locus of their

focus is ,d nh?kZ o`Ùk fudk; dk lgk;d o`Ùk ewy fcUnq ls xqtjrk gS rFkkx-v{k ,oay-v{k ij Øe'k%8 rFkk6 bdkbZ ds

vUr%[k.M dkVrk gSA ;fn ,s ls lHkh nh?kZo`Ùkks a dh mRdsUærk12 gS] rc muds ukfHk dk fcUnq iFk gSµ

(A)2 2x y

16 9 = 25 (B*) 4x 2 + 4y 2 – 32x – 24y + 75 = 0

(C)2 2x y

9 16 = 25 (D) 2x 2 + y 2 = 2

Hint.

(0, 6)

(8, 0)

Centre of ellipse (4, 3) & diameter of circle = 2a = 10distance of focus from centre = ae = 5/2

locus is (x – 4) 2 + (y – 3) 2 = 25/4 nh?kZ o`Ùk dk dsUæ (4, 3) & dsUæ ls ukfHk dh nwjh= ae = 5/2

fcUnqiFK(x – 4) 2 + (y – 3) 2 = 25/4


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9. Two tangents to the hyperbola2 2

2 2x y

–a b

= 1 having slopes m 1 & m 2 cut the coordinate axes in four con-

cyclic points. Then m 1m 2 is equal to

vfrijoy;2 2

2 2

x y –

a b = 1 dh nks Li'kZ js [kk,s a] ftudh iz o.krk,sam 1 vkS j m 2 gS] funs Z'kkad v{kks a dks pkj leo`Ùkh;

fcUnqvks a ij dkVrh gSA rcm 1m 2 dk eku gSµ

(A*) 1 (B) –1 (C)ab (D) –

ba

Hint. Let the tangents be ekuk Li'kZ js [kk,s a gSµy = m 1x +

2 2 21a m – b

y = m 2x +2 2 2

2a m – b

point of intersection of these with axes are budk v{kks a ds lkFk izfrPNsn fcUnq gS2 2 2

1

1

a m – b A – ,0

m

2 2 2

2

2

a m – bB , 0

m

2 2 21C 0, a m – b 2 2 22D 0, a m – b for concyclicity, pØh;rk ds fy, OA.OB = OC.OD m 1m 2 = 1

10. The chord of contact of a point P with respect to a hyperbola and its auxiliary circle are at right angles,then P lies on(A) conjugate hyperbola (B) directrices(C*) one of the asymptotes (D) None of these

,d fcUnqP dh ,d vfrijoy; ,oa mlds lgk;d o`Ùk ds lkis{k Li'kZ thok,sa ledks .k ij gS] rc P fLFkr gSµ (A) la ;qXeh vfrijoy; ij (B) fu;rkvksa ij(C*) ,d vUurLi'khZ ij (D) bues a ls dksbZ ugh

Hint. P(h, k)

T = 0 for hyperbola vfrijoy; ds fy, 2 2hx ky

–a b

= 1 ...(i)

T = 0 for auxilliary circle lgk;d o`Ùk ds fy, hx + ky = a 2 ...(ii)equation (i) & (ii) are perpendicular (i) vkS j (ii) yEcor~ gS

2 2

2 2

h k –

a b = 0 Asymptotes vuUrLi'khZ ;k

11. If H is the orthocentre of an acute angle triangle whose circum-circle is x 2 + y 2 = 16, thencircumdiameter of HBC is

;fn H ,d U;wudks .k f=kHkqt dk yEcdsUæ gS ftldk ifjxr o`Ùkx2 + y 2 = 16 gS] rc HBC dk ifjxr O;kl gSµ (A) 1 (B) 2 (C) 4 (D*) 8

Sol.

A

B Ca

90 – C 90 – B

H

Circum radius of ABC, HAB, HBC and HCA is same BHC ( B C) = B + C = – A

asin( A)

= 2R =a

sinA


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12. An endless inextensible string of length 15m passes around two pins, A & B which are 5m apart. Thisstring is always kept tight and a small ring, R, of negligible dimensions, inserted in this string is made tomove in a path keeping all segments RA, AB, RB tight (as mentioned earlier). The ring traces a path,given by conic C, then(A*) Conic C is an ellipse with eccentricity 1/2 (B) Conic C is an hyperbola with eccentricity 2(C) Conic C is an ellipse with eccentricity 2/3 (D) Conic C is a hyperbola with eccentricity 3/2

,d fcuk Nks j dh ¼vUrghu½ vforkU; Mks jh ftldh yEckbZ15m gS] nks fcUnqvks a (pins) A rFkkB ls xqtjrh gS] tgk¡ AB = 5m gSA ;g Mks jh ges'kk dl dj j[kh tkrh gS ,oa ,d NksVk lk oy;]R, ftldh foek,s a ux.; gS] Mksjh esa Mkyk tkrk gS tks bl izdkj xfr djrk gS fd lHkh [k.MRA, AB, RB dls (tight) gq, gS ¼tSls Åij fn;k x;k gS½A oy; ,d iFk cukrk gS og 'kkadoC gS] rc (A*) 'kkadoC ,d nh?kZ o`Ùk gS ftldh mRdsUærk1/2 gSA (B) 'kkadoC ,d vfrijoy; gS ftldh mRdsUærk 2 gSA (C) 'kkadoC ,d nh?kZ o`Ùk gS ftldh mRdsUærk2/3 gSA (D) 'kkadoC ,d vfrijoy; gS ftldh mRdsUærk 3/2 gSA

Sol.

R

AB 5Since length of string is constant, RA + RB = 10, hence locus of R, i.e. conic C is an ellipse with

eccentricity5 1

10 2.

pawfd Mksjh dh yEckbZ vpj gSRA + RB = 10 vr%R dk fcUnq iFk] nh?kZ o`ÙkC dh mRdsUærk5 110 2

.

13*. Let ABC be such that BAC =23

and AB.AC = 1, then the possible length of the angle bisector AD

is

ekuk ABC bl izdkj gS fd BAC = 23

rFkk AB.AC = 1 ] rc dks .k v}Zd AD dh laHkkfor yEckbZ gSµ

(A) 2 (B) 1 (C*) 1/2 (D*) 1/3

Sol.

60° 60°x

y

D C B

1/x

A

AD = y =2bc A

cosb c 2

y =1

1x

x

ymax. =12

14*. In a triangle ABC, If D is mid point of side BC and AD is perpendicular to AC, then the value ofcosA.cosC is ABC esa ;fnD HkqtkBC dk e/; fcUnq gS rFkk AD, AC ds yEcor~ gS rc cosA.cosC dk eku gS&

(A)22b

ac (B)

2 22 a – c3bc

(C*) –22b

ac (D*)

2 22(c – a )3ac


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Sol.

A bC

D

a/2

B

c

a/2

From ACD, ls

cosC =2ba

2 2 2a b – c 2b

2ab a 3b 2 = a 2 – c 2

Now vc , cosA . cosC =2 2 2 2 2 2 2 2b c – a 2b b c – a 2(c – a )

.2bc a ac 3ac

15. Circles are drawn with OA & OB as diameters, where A & B are points of parabola y 2 = 4x. Thesecircles meet at P (other than O). m 1 and m 2 are slope of tangents at A & B respectively and m is slopeof chord AB, then (given m 1 + m 2 0, A, B are points other than origin and 'O' is origin)

(A*) A, P, B are collinear points (B*) m is harmonic mean of m 1 and m 2 (C) m is arithmetic mean of m 1 and m 2 (D*) OP is perpendicular to ABOA o OB dks O;kl eku dj o`Ùk cuk;s tkrs gS tgk¡ A o B ijoy; y2 = 4x ij fcUnq gS ;s o`ÙkP ( ewy fcUnqO dsvykok) ij feyrs gSAm 1 ,oa m 2 Øe'k% A ,oa B ij [khaph xbZ Li'kZjs[kkvksa dh iz o.krk;sa gS rFkkm thok AB dh

izo.krk gS] rks( fn;k gSm 1 + m 2 0 o A, B ewy fcUnq ugha gSA)(A*) A, P, B la js[kh; fcUnq gSA (B*) m, m 1 rFkkm 2 dk gjkRed ek/; gSA (C) m, m 1 rFkkm 2 dk lekUrj ek/; gSA (D*) OP, AB ds yEcor~ gSA

Sol.O P

B(t2)

A(t 1)

Since OA & OB are diameters of circles OPA = OPB = 90°Hence A, P, B are collinear

Now m =1 2

2t t

=

1 2

21 1

m m

= 1 21 2

2m mm m

1 21 2

1 1m & m

t t

Hence (A), (B), (D)

Hindi.O P

B(t2)

A(t 1)

pwafdOA rFkkOB o`Ùkksa ds O;kl gSOPA = OPB = 90°


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vr% A, P, B la js [kh; gSA

vc m =1 2

2t t

=

1 2

21 1

m m

= 1 21 2

2m mm m

1 21 2

1 1m & m

t t

vr%(A), (B), (D)

16. Tangents are drawn to hyperbola

2 2

2

x y16 b = 1. ('b' being parameter) from A(0, 4). The locus of point of

contact of these tangent is a conic C, then(A*) Eccentricity of conic C is 1(B*) (0, 3) is focus of C(C) Eccentricity of conic C is 1/2(D) (0, 5) is focus of C

A(0, 4) ls vfrijoy;2 2

2x y16 b

= 1. ('b' iz kpy gS) ij Li'kZ js[kk,sa [khaph tkrh gSA bu Li'kZ js[kkvksa ds Li'kZ

fcUnqvks a dk fcUnqiFk 'kkadoC gS] rks(A*) 'kkadoC dh mRdsUærk1 gSA (B*) (0, 3), C dh ukfHk gSA (C) 'kkadoC dh mRdsUærk1/2 gSA (D) (0, 5), C dh ukfHk gSA

Sol.

(0, 4)

P(4sec , btan )

Tangent at P isx sec y tan

4 b

= 1.

It passes through (0, 4) Hence b = – 4 tan ...(1)Now h = 4 sec and k = btan = – 4 tan 2 (from (1))

K = – 4(sec 2 – 1) k = – 42h

116

.

4K – 16 = – h 2 x2 = – 4 (y – 4) (A) & (B)

Hindi.

(0, 4)

P(4sec , btan )

P ij Li'kZ js[kkxsec y tan4 b

= 1 gSA

;s (0, 4) ls xqtjrh gSA vr%b = –4 tan ...(1)vc h = 4 sec vkS j k = btan =–4 tan 2 ((1) ls)


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K = –4(sec 2 – 1) k = – 42h

116

.

4K – 16 = –h 2 x2 = –4 (y –4) (A) rFkk(B)

17. From centre O, of the ellipse2 2x y

16 9 = 1, two perpendicular rays are drawn meeting the ellipse at P &

Q, N is the foot of perpendicular from O to PQ, then

nh?kZ o`Ùk2 2x y

16 9 = 1 ds dsUæ O ls nks yEcor~ fdj.ks a nh?kZ o`Ùk dksP ,oa Q ij feyrh gSAN, PQ ij O ls yEc ikn

gS] rks

(A*)2 2

1 1 25144OP OQ

(B)2 2

1 1 25144OP OQ

(C*) ON =125

(D) ON =65

Sol. Let OP = r 1 & OQ = r 2

P(r 1cos ,r 1sin )

O

N

Now P & Q lie on the ellipse hence

r 12

2 2cos sin16 9

= 1

2 2cos sin16 9

= 2

1

1

r ….(1)

r 22 2 2sin cos

16 9

= 1 2 2sin cos16 9

= 22

1r

…..(2)

Now (1) + (2) 2 21 2

1 1 1 116 9 r r

= 25144

Let equation of chord PQ be x cos + ysin = p, homogenizing the equation of ellipse with this chordgives

2 2x y16 9

–2

xcos y sinp

= 0

As OP & OQ are perpendicularcoefficient of x 2 + coefficient of y 2 = 0

2

21 cos

16 p

+2

21 sin9 p

= 0 21 1 116 9 p p2 = 144

25 p = 12/5

Hindi. ekukOP = r 1 rFkkOQ = r 2


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P(r,cos ,r,sin )

vc P rFkkQ nh?kZ o`Ùk ij gS vr%

r 12

2 2cos sin16 9

= 1

2 2cos sin16 9

= 2

1

1

r ….(1)

r 22

2 2sin cos16 9

= 1

2 2sin cos16 9

= 2

2

1

r …..(2)

vc (1) + (2) 2 21 2

1 1 1 116 9 r r

= 25144

ekukPQ thok dk lehdj.k gSx cos + ysin = p nh?kZ o`Ùk ds lehdj.k dks le?kkr cukus ij 2 2

x y16 9 –

2

xcos y sinp = 0

D;ks afdOP rFkkOQ yEcor~ gSA x2 dk xq .kkad+ y 2 dk xq .kkad= 0

2

21 cos

16 p

+2

21 sin9 p

= 0 21 1 1

16 9 p

p 2 =14425

p = 12/5

18. y = x is tangent to an ellipse whose foci are (1, 0) and (3, 0) then(A) Major axis of ellipse is = 6

(B*) Major axis of ellipse is = 10

(C*)3 3

,4 4

is the point of contact of this ellipse and this tangent

(D)1 1

,2 2

is the point of contact of this ellipse and this tangent

js [kky = x nh?kZo`Ùk] ftldh ukfHk;k¡(1, 0) ,oa (3, 0) gSa] dh Li'kZ js [kk gS] rks(A) nh?kZ o`Ùk dk nh?kZv{k= 6 gSA (B*) nh?kZ o`Ùk dk nh?kZv{k = 10 gSA

(C*)3 3

,4 4

nh?kZ o`Ùk vkS j Li'kZ js [kk dk Li'kZ fcUnq gSA

(D)1 1

,2 2

nh?kZ o`Ùk vkS j Li'kZ js [kk dk Li'kZ fcUnq gSA

Sol. Product of perpendicular from two foci on any tangent = b 2 =3 1

.2 2

=32

b =32

Now ae = 1 a 2 = b 2 + a 2e 2 a =52

We know that tangent and normal bisect the angle between focal distances of a point.


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y = x

(3, 0)(1, 0)

(0, 1)

Image of (1, 0) in y = x is (0, 1), line joining (0, 1) & (3, 0) is x + 3y = 3. Point of contact of y = x & ellipse

is the point of intersection of y = x and x + 3y = 3, i.e.3 3

,4 4

Hindi. ukfHk;ksa ls Li'kZ js[kkkvksa ij Mkys x;s yEcksa dk xq.kuQy= b 2 = 3 1.2 2

=32

b =32

vc ae = 1 a 2 = b 2 + a 2e 2 a = 52

ge tkurs gS fd Li'kZ js[kk ,oa vfHkyEc ukfHk;ksa nw jh;ks a ds e/; cuus okys dks a .k dks lef}Hkkftr djrs gSA

y = x

(3, 0)(1, 0)

(0, 1)

(1, 0) dk js[kky = x esa izfrfcEc(0, 1) gSA fcUnqvksa(0, 1) rFkk(3, 0) dks feykus okyh js[kkx + 3y = 3 gSA

y = x dk Li'kZ fcUnq rFkk js [kky = x vkS j x + 3y = 3 dk iz frPNsn fcUnq gS vFkkZr~ 3 3,4 4

gSA

19. Let set S consists of all the points (x, y) satisfying 9x 2 + 16y 2 144. For points in S let maximum and

minimum value ofy 4x 9

be M and m respectively, then

Ekkuk leqPp;S mu lHkh fcUnqvks a (x, y) dk leqPp; gS tks 9x 2 + 16y 2 144 dks larq"V djrs gSA leqPp;S es a

fLFkr fcUnqvksa (x, y) ds fy;s O;atd y 4x 9

dk vf/kdre ,oa U;w ure eku Øe'k%M rFkkm gS] rks

(A*) M = 1 (B) M =657

(C) m = 1 (D*) m =7

65

Sol.

A(9, 4)

y 4x 9

is the slope of line joining A(9, 4) & (x, y)

For maximum & minimum value of this expression we have to determine the slope of tangents to the

ellipse2 2x y

25 16

= 1 from (9, 4)


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Hence y = Kx ± 216k 9 It passes through (9, 4)Hence (4 – 9k) 2 = 16K 2 + 9 65K 2 – 72K + 7 = 0

Hence K = 1 or7

65 M = 1 & m =

765

Hindi.

A(9, 4)

A(9, 4) vkS j (x, y) dks tksM+us okyh js[kk dh izo.krky 4x 9

gS

vf/kdre vkS j U;wure eku fudkyus ds fy, gesa Li'kZ js[kk dh izo.krk fudkyuh gksxhAvr%y = Kx ± 216k 9

;g (9, 4) ls xqtjrh gSA

vr%(4 – 9k)2

= 16K2

+ 9 65K2

– 72K + 7 = 0vr% K = 1 ;k 7

65 M = 1 rFkkm = 7

65

20. Consider the curve ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0, where x, y are real variables and a, b, c , f, g, hare real constants. Let = abc + 2fgh – af 2 – bg 2 – ch 2, and curve S be the locus of point of intersectionof perpendicular tangents of the above curve.(A*) If 0 and h 2 = ab, then S is a straight line

(B) If 0, h = 0, a = b 0 then S is a circle of radius 2 22 g f c (C*) If = 0, a + b = 0, then S is a point only(D) IF = 0, a + b = 0 then S is a pair of straight lines.

oØ ax2

+ 2hxy + by2

+ 2gx + 2fy + c = 0 ij fopkj dhft;s tgk¡ x, y okLrfod pj gS ,oaa, b, c , f, g, h okLrfod vpj gSA ekuk = abc + 2fgh – af 2 – bg 2 – ch 2 ,oa oØ S, fn;s x, oØ dh yEcor~ Li'kZ js[kkvksa ds izfrPNsn fcUnqvksa dk fcUnqiFk gS&(A*) S ,d ljy js[kk gS ;fn 0 vkS j h2 = ab.

(B) S ,d o`Ùk gS ftldh f=kT;k 2 22 g f c gS ;fn 0, h = 0, a = b 0.(C*) S dsoy ,d fcUnq gksxk ;fn = 0, a + b = 0.(D) S ljy js[kk ;qXe gksxk ;fn = 0, a + b = 0.

Sol. If 0, h 2 = ab curve is a parabola, hence S is a straight line

If 0, h = 0, a = b 0 curve is a circle & S is a circle of radius 2 22 g f c (provided a = b = 1)If = 0, a + b = 0 curve is a pair of perpendicular straight lines for which S is a point which is thepoint of intersection of the two lines.

Hindi. ;fn 0, h 2 = ab oØ ijoy; gS vr% S ljy js [kk gSA

;fn 0, h = 0, a = b 0 oØ o`Ùk gS rFkkS dh f=kT;k 2 22 g f c gS ( ;fn a = b = 1) ;fn = 0, a + b = 0 oØ yEcor~ js [kk ;qXe gksxk ftlds fy,S ,d fcUnq gksxkA


13/27




21. The ellipse2x

4 +

2y3

= 1 has a double contact with a circle at the extremity of latus rectum. The point

of contact lying in first and fourth quadrant.(A) Centre of circle is (0, 0)

(B*) Centre of circle is1

,04

(C*) Radius of circle is3 5

4

(D) Radius of circle is3 5

2

nh?kZ o`Ùk2x

4 +

2y3

= 1 ,d o`Ùk dks ukfHk yEc ds nksuks fljs fcUnqvksa ij Li'kZ djrk gSA Li'kZ fcUnq izFke ,oa prqFkZ

prqFkkZ± 'k es a gS] rks(A) o`Ùk dk dsUæ(0, 0) gSA

(B*) o`Ùk dk dsUæ 1 ,04

gSA

(C*) o`Ùk dh f=kT;k3 54

gSA

(D) o`Ùk dh f=kT;k 3 52

gSA

Sol.C

3P 1,

2

31,

2

By symmetry centre of circle lies on x-axis

Normal at P is4x 3y1 3 / 2

= 1 point C is1

,04

Radius =2 2

1 31

4 2

=9 9

16 4 =

3 54

Hindi.C

3P 1,

2

31,

2

lefefr ls o`Ùk dk dsUæx-v{k ij gksxkA

P ij vfHkyEc 4x 3y1 3 / 2

= 1 gS fcUnqC 1 ,04

gSA

f=kT;kk=2 2

1 31

4 2

= 9 916 4

=3 5

4


14/27




22. Normal at point P(x 1, y 1), not lying on x-axis, to the hyperbola x2 – y 2 = a 2 meets x-axis at A and y-axis

at B. If O is origin then(A*) Circumcentre of triangle OAB is P.(B*) Slope of OP + slope of AB = 0(C) Slope of OP = slope of AB(D*) Locus of centroid of triangle OAB is a rectangular hyperbolavfrijoy; x2 – y 2 = a 2 ds fcUnqP(x 1, y 1), tks fd x-v{k ij ugha gS] ij vfHkyEc [khapk tkrk gS] tks x-v{k ,oay-v{k dks Øe'k% A vkS j B ij feyrk gS] O ewy fcUnq gS] rc(A*) f=kHkqtOAB dk ifjdsUæP gSA (B*) OP dh izo.krk + AB dh iz o.krk = 0(C) OP dh iz o.krk = AB dh iz o.krk (D*) f=kHkqtOAB ds dsUæd dk fcUnqiFk vk;rh; vfrijoy; gSA

Sol. Equation of normal at P is1 1

x yx y

= 2 A(2x 1, 0), B(0, 2y 1)

Hence P is mid-point of AB, i.e. circumcentre of OAB

m AB = – 11

yx

, m OP = 11

yx

Let (h, k) be centroid of the triangle OAB 3h = 2asec 1 3k = 2atan

x2 – y 2 =24a

9

Hindi. P ij vfHkyEc dk lehdj.k1 1

x yx y

= 2 gS A(2x 1, 0), B(0, 2y 1)

vr%P, AB dk e/; fcUnq gS vFkkZ r~ f=kHkqtOAB dk ifjdsUæ

m AB = – 11

yx

, m OP = 11

yx

ekuk(h, k) f=kHkqtOAB dk dsUæd gSA 3h = 2asec 1 3k = 2atan

x2

– y2

=

24a9

23. A & B two points on the curve xy = a 2. Let N be the mid-point of AB. The line through A and B meets.x-axis at P and y-axis at Q, then(A*) N bisects PQ(B) ON is perpendicular to AB (where O is origin)(C*) AP = BQ(D*) AQ = BP

A ,oa B oØ xy = a 2 ij nks fcUnq gSA ekukN, AB dk e/; fcUnq gSA fcUnqvksa A rFkkB ls xqtjus okyh ljy js[kkx-v{k ,oay-v{k dks Øe'k%P o Q ij feyrh gS] rks (A*) N, PQ dks lef}Hkkftr djrk gSA (B) ON, AB ds yEcor~ gS ( tgk¡ O ewy fcUnq gS)(C*) AP = BQ(D*) AQ = BP

Sol.

Q B

A

N

P

22

a(at , )

t

21

a(at , )

t

Equation of AB is x + t 1t2y = a(t 1 + t 2)


15/27




Hence point P is (a(t 1 + t 2), 0) and Q is1 2

1 10,a

t t

N is 1 2

1 2

a t t a 1 1,

2 2 t t

Hence N bisects AB as well as PQ

m ON = 1 2

1 2

a t t

2t t.

1 22

a t t =

1 2

1

t t

Now AN = BN and PN = QN AP + AN = BQ + BN AP = BQ

Further AP + AB = BQ + AB BP = AQ

Hindi.

Q B

A

N

P

AB dk lehdj.k x + t 1t2y = a(t 1 + t 2) gSA

vr% fcUnqP (a(t 1 + t 2), 0) vkS j Q1 2

1 10,a

t t

gSA

N 1 2

1 2

a t t a 1 1,

2 2 t t

gSA

vr%N, AB dks lef}Hkkftr djrk gS lkFk ghPQ dks Hkh

m ON = 1 2

1 2

a t t

2t t.

1 22

a t t =

1 2

1t t

vc AN = BN vkS j PN = QN AP + AN = BQ + BN AP = BQ blhfy, AP + AB = BQ + AB BP = AQ

24. Let, S, be a conic whose centre is M(p, q). Locus of middle points of chords of this conic, which passesthrough a fixed point N( , ) is(A*) Another conic which has a centre (B) Another conic with same focus

(C*) Another conic with centre asp q

,2 2

(D) Another conic with centre asp q

,2 2

ekuk, S ,d 'kkado gS ftldk dsUnzM(p, q) gS rks 'kkado dh lHkh thokvksa ds e/; fcUnq dk fcUnqiFk] tks ,d fuf'pr fcUnqN( , ) ls xqtjrh gSa] gksxk (A*)

,d nw ljk 'kka do ftldk dsUæ gSA(B)

,d nwljk 'kkado leku ukfHk okyk gksA(C*) ,d nw ljk 'kkado ftldk dsUæ p q,

2 2

gSA (D) ,d nw ljk 'kkado ftldk dsUæ p q,

2 2

gSA

Sol. Since the above conic has a centre it must be a hyperbola or an ellipseLet origin be shifted to M(p,q) and axis be so rotated that it coincides with the principle axis of conic S,hence its equation is Ax 2 + By 2 = 1, and new-co-ordinates of N be ( ', ')Equation of chord whose mid-point is (h, k) is T = S 1, i.e.

Axh + Byk = Ah 2 + Bk 2 it passes through ( ', ')Hence A(x 2 – x ') + B(y 2 – y ') = 0


16/27




A2

'x

2

+ B2

'y

2

= 2 A '

4 +

2B '4

Hence locus is a similar conic whose centre is' ',

2 2

i.e. mid-point of MN.Hindi. pawfd mijksDr 'kkado dk dsUæ gS vr% ;s nh?kZ o`Ùk ;k vfrijoy; gksxkA

ekuk ewy fcUnq dksM(p,q) ij LFkkukUrfjr djrs gS rFkk v{k dks 'kkadoS ds laikrh djrs gq, /kw .kZu djrs gSA vr% bldk lehdj.k Ax 2 + By 2 = 1 vkSjN ds u;s funsZ'kkad( ', ') gSA thok dk lehdj.k ftldk e/; fcUnq (h, k) gS T = S 1, i.e. Axh + Byk = Ah 2 + Bk 2 ;g ( ', ') ls xqtjrh gSA vr% A(x 2 – x ') + B(y 2 – y ') = 0

A2

'x

2

+ B2

'y

2

= 2 A '

4 +

2B '4

vr% le:i 'kkado dk fcUnqiFk ftldk dsUæ ' ',2 2

gSA

vFkkZ r~ e/; fcUnq MN gSA

25. Consider the ellipse2 2

2x y

f(k 11)f(k 2k 5) = 1, where f(x) is a strictly decreasing positive function,

then(A*) the set of values of k for which the major axis of the ellipse is x-axis is (–3, 2)(B) the set of values of k for which the major axis of the ellipse is y-axis is (– , 2)(C*) the set of values of k for which the major axis of the ellipse is y-axis is (– , –3) (2, )(D) the set of values of k for which the major axis of the ellipse is x-axis is (–3, )

ekuk fd ,d nh?kZo`Ùk2 2

2x y

f(k 11)f(k 2k 5) = 1 gS] tgk¡ f(x) fujUrj ãleku /kukRed Qyu gS] rc

(A*) ;fn nh?kZ o`Ùk dk nh?kZv{k

x-v{k gS] rc

k ds ekuks a dk leq Pp;

(–3, 2) gSA

(B) ;fn nh?kZ o`Ùk dk nh?kZv{k y-v{k gS] rck ds ekuksa dk leq Pp; (– , 2) gSA (C*) ;fn nh?kZ o`Ùk dk nh?kZv{k y-v{k gS] rck ds ekuksa dk leq Pp; (– , –3) (2, ) gSA (D) ;fn nh?kZ o`Ùk dk nh?kZv{k x-v{k gS] rck ds ekuks a dk leqPp; (–3, ) gSA

Hint. For major axis to be x-axis, nh?kZ v{k ds fy, x-v{k gSA f(k2 + 2k + 5) > f(k + 11) k2 + 2k + 5 < k + 11 k (–3, 2)

26. Two concentric ellipses are such that the foci of one lie on the other and the length of their major-axesare equal. If e 1 & e 2 be their eccentricities, then (A*) the quadrilateral formed by joining their foci is a parallelogram

(B*) the angle between their axes is given by cos =2 2 2 21 2 1 2

1 1 1 –

e e e e

(C*) their axes are perpendicular if e 1 =221– e

(D) None of these

nks ladsUæh; nh?kZ o`Ùk bl izdkj gS fd ,d dh ukfHk;k¡ nwljs ij fLFkr gS rFkk muds nh?kZ v{kksa dh yEckbZ cjkA ;fn mudh mRdsUærk,s ae 1 rFkke 2 gS] rc(A*) muds ukfHk;ksa dks feykus ls cuk prqHkq Zt ,d lekUrj prq HkqZt gSA

(B*) muds v{kks a ds e/; dks .k cos = 2 2 2 21 2 1 2

1 1 1 –

e e e e

(C*) mudh v{k yEcor~ gS ;fne 1 = 221– e


17/27




(D) bues a ls dksbZ ugh

Hint.

C2H

o SS¢H¢

HH¢ & SS ¢ have same mid-point HSH ¢S ¢ is a parallelogram

let one ellipse be2 2

2 2

x ya b

= 1 ...(i) H lies on it

also H (ae 2 cos , ae 2 sin )putting in equation (i)

cos 2 = 2 2 2 21 2 1 2

1 1 1

e e e e

(A), (B) & (C) are correct. (C) follows from (B)

Hint.

C2Ho SS¢

H¢

HH¢ rFkkSS ¢ leku fcUnq j[krs gS HSH ¢S ¢ lekUrj prqHkqZt gS

ekuk ,d nh?kZ o`Ùk2 2

2 2x ya b

= 1 gS ...(i) H bl ij fLFkr gS

rFkkH (ae 2 cos , ae 2 sin ) lehdj.k (i) esa j[kus ij

cos 2 = 2 2 2 21 2 1 21 1 1

e e e e (A), (B) & (C) are correct. (C) follows from (B)

27_. An ellipse2 2

2 2x ya b

= 1 (a > b) contains a circle (x – 1) 2 + y 2 = 1 so that the area of ellipse is minimum,

then

(A*) a 2 + b 2 = 6 (B*) a 2 – b 2 = 3

(C) Min. Area of ellipse =32

sq. units (D*) Minimum area of ellipse =3 3

2 sq. units

,d nh?kZ2 2

2 2

x y

a b = 1 (a > b) o`Ùkks a (x – 1) 2 + y 2 = 1 dks bl izdkj j[krk gS fd nh?kZ o`Ùk dk {ks=kQy U;wure gSA

(A*) a 2 + b 2 = 6 (B*) a 2 – b 2 = 3

(C) nh?kZo`Ùk dk U;w ure {ks=kQy= 32

oxZ bdkbZ (D*) ) nh?kZ o`Ùk dk U;wure {ks=kQy = 3 32

oxZ bdkbZ gSA

Sol. (1, 0)

y

(0, 0) x

Solve equation of circle & ellipse


18/27




2 2

2 2x (1 (x 1) )

1a b

(b 2 – a 2)x2 + (2a 2)x – a 2b2 = 0 .......(1)

Cricle and ellipse touches each other for minimum area We get repeated roots of equation (1)

D = 0 4a 4 + 4a 2b2(b 2 – a 2) = 0 a 2 =4

2b

b 1

Area of ellipse = ab

Let A = 2a 2b2 =62

2b

b 1

to minimum area of ellipse, we have to minimum A

dA

0db

b =32

for b =32

,9

a2

Minimum area of ellipse = 9a2

× 3 3 32 2

sq. units

Hind i

(1, 0)

y

(0, 0) x

o`Ùk vkS j nh?kZ o`Ùk ds lehdj.k dks gy djus ij

2 2

2 2

x (1 (x 1) )1

a b

(b 2 – a 2)x2 + (2a 2)x – a 2b2 = 0 .......(1)

U;wure {ks=kQy ds fy, o`Ùk vkS j nh?kZ o`Ùk ,d nq ljs dks Li'kZ djrs gSA lHkh(1) ds ewy iqujko`fÙk gSA

D = 0 4a4

+ 4a2

b2

(b2

– a2

) = 0 a2

=

4

2

bb 1

nh?kZ o`Ùk dk {ks=kQYk= ab

ekuk A = 2a 2b2 =6

22

b

b 1

nh?kZ o`Ùk ds U;wure {ks=kQy ds fy, yEckbZ A

dA

0db

b =32

b =32

,9

a2 ds fy,

nh?kZ o`Ùk dk U;wure {ks=kQYk = 9a 2 × 3 3 32 2 oxZ bdkbZ

28_. Equation for an ellipse is2 2x y

13 2

. line BD passes through focus F 1 and intersects the ellipse at

points B and D. Line AC passes through focus F 2 and intersects the ellipse at points A and C. Line BDis perpendicular to AC at point P. Which of the following is (are) correct for area of quadrilateral to beminimum.(A*) The two lines AC & BD intersects on minor axis

(B*) Minimum area of quadrilateral is9625

sq. units.

(C*) circumcentre of PF 1 F 2 is centre of ellipse.


19/27




(D) Area of quadrilateral is2596

sq. units.

ekuk fd nh?kZo`Ùk dk lehdj.k2 2x y

13 2

gS js[kkBD ukfHkF1 ls xqtjrh gS rFkk nh?kZo`Ùk dks fcUnqvksaB rFkkD.

ij izfrPNs n djrh gSA js[kk AC ukfHkF2 ls xqtjrh gS rFkk nh?kZo`Ùk dks fcUnqvksa A rFkkC ij izfrPNsn djrh gSA js [kkBD, AC ds yEcor~P ij gS rks prqHkZqt dk {ks=kQYk U;wure gks us ds fy, fuEu es a ls dkSulk lgh gS\(A*) nks js[kk,s a AC rFkkBD y?kq v{k ij izfrPNsn djrh gSA

(B*) prqHkq Zt dk U;w ure {ks=kQy9625

oxZ bdkbZ gSA

(C*) PF 1 F 2 i dk ifjdsUnz nh?kZ o`Ùk dk dsUnz gSA

(D) prqHkqZt dk {ks=kQy 2596

oxZ bdkbZ gSA

Sol.

B P

90°+ F 1

(1,0)

y

xDF 2

(–1,0)

A

C

Parametric coordinates of line passing through F 1(1, 0) is Q(1 + rcos(90° + ), r sin(90° + ) Q(1 – rsin , r cos ) This point lies on ellipse

2 2(1 r sin ) (r cos )

13 2

(2sin 2 + 3cos 2 )r 2 – (4sin )r – 4 = 0

(2 + cos 2 )r 2 – (4sin ) r – 4 = 0It is quadratic in r, which will gives points. B & D.

|r 1 – r

2| = BD =

2

2 2 2 2

16sin 4 4 3

(2 cos ) (2 cos ) 2 cos

similarly AC =2

4 32 sin

Area of quadrilateral ABCD =12

(BD) × (AC) =2 2

1 4 3 4 32 2 cos 2 sin

=2 2

24

6 sin cos = 2

24

sin 26

4

For area to be minimum, sin 22 = 1 =4

Minimum area of quadrilateral ABCD =9625

sq. units.

slope of line BD is –1 and slope of line AC is 1equation of line BD is y = –x + 1 & that of line AC is y = x + 1Point of intersection lines BD & AC is (0, 1), which lies on minor axis.


20/27




Hindi.

B P

90°+ F 1

(1,0)

y

xDF 2(–1,0)

A

C

F 1(1, 0) ls xqtjus okyh js [kk ds izkpfyd funsZ 'kkadQ(1 + rcos(90° + ), r sin(90° + ) gS Q(1 – rsin , r cos ) ;g fcUnq nh?kZ o`Ùk ij fLFkr gS

2 2(1 r sin ) (r cos )

13 2

(2sin 2 + 3cos 2 )r 2 – (4sin )r – 4 = 0

(2 + cos 2 )r 2 – (4sin ) r – 4 = 0 ;g r, esa f}?kkr gS tksB rFkkD fcUnqvksa ij gSA

|r 1 – r 2 | = BD =2

2 2 2 216sin 4 4 3

(2 cos ) (2 cos ) 2 cos

bl izdkj AC = 24 3

2 sin

prqHkqZt dk {ks=kQYk ABCD = 12

(BD) × (AC) =2 2

1 4 3 4 32 2 cos 2 sin

=2 2

24

6 sin cos = 2

24

sin 26

4

U;wure gks us ds fy,, sin 22 = 1 =4

prqHkqZt ABCD dk U;w ure {ks=kQy= 9625

oxZ bdkbZ

js [kkBD dh iz o.krk –1 gS rFkk js[kk AC dh iz o.krk 1 gSA js [kkBD dk lehdj.k y = –x + 1 rFkk js [kk AC dk lehdj.k y = x + 1BD rFkk AC js [kkvks a ds izfrPNsn fcUnq(0, 1), gS tks fd y?kqv{k ij gSA

29. A tangent drawn to the hyperbola2

2xa

–2

2yb

= 1 at P6

forms a triangle of area 3a 2 sq. units. with co-

ordinate axes. If the eccenticity of the hyperbola is 'e', then the value of e 2 – 9 is.

vfrijoy;2

2x

a

–2

2y

b

= 1 ds fcUnqP6

ij Li'kZ js[kk [khaph tkrh gS tks funsZ'kkad v{kks ds lkFk3a 2 oxZ bdkbZ

{ks=kQy dk f=kHkq t cukrh gSA ;fn vfrijoy; dh mRdsUærk'e' gS rc e 2 – 9 dk eku gSµAns. 8

Sol. P asec ,b tan6 6

2a b

,3 3

Tangent at P ij Li'kZ js [kk 2x3a

–y

3b = 1

Area {ks=kQy = 3a 2 = 12

.3a2

. 3 b

b

a

= 4


21/27




30. AB is focal chord of a parabola. Let D and C be foot of perpendicular from A & B on it's directrixrespectively. If CD= 6 and area of trapezium ABCD is 24 square units, then find length of chord AB.

AB ijoy; dh ukfHk; thok gSA ekukD rFkkC Øe'k% A rFkkB ls ijoy; dh fu;rk ij yEcikn gSA ;fn CD = 6 bdkbZ ,oa leyEc prqHkqZt ABCD dk {kS=kQy24 oxZ bdkbZ gS] rks AB dh yEckbZ Kkr dhft;sA

Ans. 8

Sol.

CB

S

AD

6

Let S be focus AS =AD & BS = BC Area of trapezium

=21

{AD + BC}.6

= 3 (AS + BS)= 3ABhence AB = 8 units

Hindi.

CB

S

AD

6

ekukS ukfHk gS AS =AD rFkkBS = BC leyEc prqHkq Zt dk {kS=kQy

=21

{AD + BC}.6

= 3 (AS + BS)= 3ABvr% AB = 8 bdkbZ

31. A circle is drawn whose centre is on x-axis and it touches y-axis. If no part of the circle lies ouside theparabola y 2 = 8x, then maximum possible radius of the circle is

,d o`Ùk dk dsUnzx- v{k ij gS ,oa ;g o`Ùky-v{k dks Li'kZ djrk gSA ;fn o`Ùk dk dksbZ Hkh Hkkx ijoy;y2 = 8x ds

ckgj ugha gS] rks o`Ùk dh vf/kdre laHkkfor f=kT;k gSµ Ans. 4

Sol.

Let equation of circle be (x – r) 2 + y 2 = r 2 Solving at with y 2 = 8x 1 we getx2 – 2rx + r 2 + 8x = r 2 x = 0 or x = 2r – 8

Now 2r – 8 0 r 4 Hence r max = 4


22/27




Alter: Normal at (2t 2, 4t) to y 2 = 8x, meets x-axisat (4 + 2t 2, 0), So x-coordinate of centre should be such that r 4 + 2t 2 4, hence r max = 4

Hindi.

ekuk o`Ùk(x – r) 2 + y 2 = r 2 gSA y2 = 8x 1 ds lkFk gy djus a ijx2 – 2rx + r 2 + 8x = r 2 x = 0 or x = 2r – 8vc 2r – 8 0 r 4 vr%r max = 4

Alter: vfHkyEc(2t 2, 4t) ij ijoy; y2 = 8x dk x-v{k dks(4 + 2t 2, 0) ij feyrk gSA vr% dsUæ ds funsZ 'kkad bl iz dkj gksaxsar 4 + 2t 2 4, vr%r max = 4

32. Parabola, P 1 has focus at S(2, 2) and y-axis is it's directrix. Parabola, P 2 is confocal with P 1 and it'sdirectrix is x-axis. Let Q(x 1, y 1) and R(x 2, y 2) be real points of intersection of parabolas P 1 and P 2.

If the ratioRS

a b bQS

find (a + b) (given x 2 > x 1 and a, b N)

ijoy; P 1 dh ukfHkS(2, 2) o fu;rk y-v{k gS ijoy; P 2 ijoy; P 1 ds lkFk lgukfHk; gS rFkk bldh fu;rkx-

v{k gS ekukQ(x 1, y 1) o R(x 2, y 2), P 1 o P 2 ds okLrfod iz frPNsn fcUnq gS aA ;fn RS a b bQS

rks (a + b) gS

( fn;k gSx2 > x 1 o a, b N)Ans. 5Sol.

R

P 2

S(2, 2)

Q

P 1

21

22

P (y 2) 4(x 1)

P (x 2) 4(y 1)

Subtracting them we get (x – y)(x + y) = 0 line QR is y = x

Hence x 1 & x2 are roots of the equation

(x – 2) 2 = 4(x – 1) x2 – 8x + 8 = 0 1

2

4 2 2 x

4 2 2 x

(given x 2 > x 1)

SoRS 2 2 2 2 1

3 2 2QS 2 2 2 2 1

Hindi.


23/27




R

P 2

S(2, 2)

Q

P 1

21

22

P (y 2) 4(x 1)

P (x 2) 4(y 1)

?kVkusa ij(x – y)(x + y) = 0 js[kkQR, y = x gS

vr% lehdj.k ds ewyx1 rFkkx2 gSA are roots of the equation

(x – 2) 2 = 4(x – 1) x2 – 8x + 8 = 0 1

2

4 2 2 x

4 2 2 x

( fn;k gSx2 > x 1)

vr% RS 2 2 2 2 1 3 2 2QS 2 2 2 2 1

33_. Consider a circle C : x 2 + y 2 – 8y + 12 = 0 and an ellipse E :2 2

2 2x ya b

= 1 (a > b and b < 2).

If the maximum perpendicular distance from the foci of the ellipse upon the tangent drawn to the circle

is 7 units, and shortest distance between both the curves is 1 unit, then find the value of(a 2 – 2b 2).

ekuk fd o`ÙkC : x 2 + y 2 – 8y + 12 = 0 rFkk nh?kZ o`ÙkE :2 2

2 2

x ya b

= 1 (a > b vkS j b < 2).

;fn nh?kZo`Ùk dh ukfHk;ksa ls] o`Ùk dh Li'kZ js[kk ij vf/kdre yEcor~ nw jh 7 bdkbZ gS rFkk nks oØksa ds e/; y?kqÙke nw jh1 bdkbZ gSA rc(a 2 – 2b 2) dk eku Kkr dhft,A

Ans. 8Sol. Shortest distance ¼y?kqÙke nq jh½= 1 2 – b b = 1

SM = 7 SC = 5

x

y

M 6

C(0,4)

S(ae,0)O

1

S¢

2

A

2 2a e 16 = 5 a 2e 2 = 9 a 2 – b 2 = 9 a 2 = 10 a 2 – 2b 2 = 8


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P(h, k)

P(h, k)

Comprehension (Q. No. 34 to 35)_vuqPNsn(iz u la[;k34 ls 35)

From a point 'P' three co-normals to the parabola y 2 = 4x such that the product of tangents of anglemade by two of them is 2. Suppose the locus of the point 'P' is a part of a conic 'C'. Now a circle S = 0is described on the chord of the conic 'C' as diameter passing through the point (1, 0) and with gradientunity, then answer the following questions :

fcUnq'P' ls ijoy; y2 = 4x ds rhu lg vfHkyECk fcUnq] bl iz dkj gS fd mu esa ls nks ds lkFk cuk, x, dks.k dh

Li'kZ T;k dk xq .kuQy 2 gS ekuk fd fcUnqP dk fcUnqiFk] 'kka do C dk ,d Hkkx gS vc 'ka kdo C dh thok dks O;kl ekudj [khpk x;k o`ÙkS = 0 ] fcUnq(1, 0) ls xqtjrk gS rFkk izo.krk bdkbZ gS rc fuEu iz 'uksa ds mÙkj nhft,A

34_. Locus of 'P' is :(A) circle (B*) Parabola (C) Ellipse (D) Hyperbola'P' dk fcUnqiFk gS:(A) o`Ùk (B*) ijoy; (C) nh?kZo`Ùk (D) vfrijoy;

35_. Radius of circle S = 0 is : o`ÙkS = 0 dh f=kT;k gSA & (A*) 4 (B) 5 (C) 17 (D) 23

Sol. Equation of normal in y = mx – 2m – m 3 it passes through P(h, k)

m 3 + (2 –h)m + k = 0

m 1m 2m 3

.......(i)

m 1 m 2 m 3 = –k, but m 1m 2 = 2

m 3 =k

2, it satisfies equation (1)

3k

8 – (2 – k)

k2

+ k = 0

k3 + 4(2 – h)k – 8k = 0 (k 0)

k2

+ 8 – 4h – 8 = 0 locus of 'P' is y 2 = 4x which a parabola.Now chord passing through (1, 0) is the focal chord.Given that slope of focal chord is 1.

1 2

2t t

= 1 t1 + t 2 = 1 and t 1t2 = –1 .....(ii)

Equation of circle described on t 1, t 2 as diameter is 2 21 2 1 2(x t )(x t ) (y 2t )(y 2t ) 0 On simplifying and using (2), we get

x2 + y 2 – 6x – 4y – 3 = 0 Radius = 4

Hindi. vfHkyEc dk lehdj.ky = mx – 2m – m 3 gSA ;g P(h, k) ls xqtjrk gS

m 3 + (2 –h)m + k = 0

m 1m 2m 3

.......(i)

m 1 m 2 m 3 = –k, but m 1m 2 = 2

m 3 =k

2, lehdj.k (i) dks larq"B djrk gSA

3k

8 – (2 – k)

k2

+ k = 0

k3 + 4(2 – h)k – 8k = 0 (k 0) k2 + 8 – 4h – 8 = 0


25/27




'P' dk fcUnqiFky2 = 4x gS tksfd ojoy; gS ;g thok (1, 0) ls xqtjrh gS tks ukfHk; thok gSA fn;k x;k gS fd ukfHk; thok dh iz o.krk 1 gSA

1 2

2t t

= 1 t1 + t 2 = 1 rFkk t1t2 = –1 .....(ii)

t1, o t2 dk O;kl ekudj [khpsa x, o`Ùk dk lehdj.k 2 21 2 1 2(x t )(x t ) (y 2t )(y 2t ) 0 (2) dh lgk;rk ls ljy djus ij

x2 + y 2 – 6x – 4y – 3 = 0 f=kT;k= 4

Comprehension (Q. No. 36 to 38)vuqPNsn(iz u la[;k36 ls 38)

The triangle ABC is inscribed in a circle of unit radius. If A : B : C = 1 : 2 : 4, then bdkbZ f=kT;k ds o`Ùk ds vUrxZr ABC gS ;fn A : B : C = 1 : 2 : 4 rc

36. cos2A + cos2B + cos2C =

(A)12

(B) –1 (C*) –12

(D) –13

37. a 2 + b 2 + c 2 =

(A)72

(B*) 7 (C) 14 (D)152

38. The area of ABC is ABC dk {ks=kQy gS&

(A) 7 (B) 7 (C)7

2 (D*)

74

Sol. A =7

, B = 27

, C = 47

A

B Ca

bc

o

1

11

(36) cos2A + cos2B + cos2C = – 1 – 4cosA cosB cosC

= – 1 – 4 cos7

cos 27

cos 47

= – 1 – 4

8sin

17 –

28sin7

(36) cos2A + cos2B + cos2C =1

–2

2 2 21 1– a 1 1– b 1 1– c 1 –

2.1.1 2.1.1 2.1.1 2 a 2 + b 2 + c 2 = 7


26/27




(38) =12

(sin2A + sin2B + sin2C) = 2sinA sinB sinC

= 2.sin7

sin27

sin47

= 2sin7

sin27

sin37

= 2. 2 2 22 3

sin .sin sin7 7 7

= 2.

7–1

72

=7

4square units oxZ bdkbZ

Comprehension (Q. 39 to 40)

Consider the circle, S, with equation x 2 + y 2 + 2gx + 2fy + c = 0. This circle meets the parabolay2 = 4ax at A(x 1, y 1), B(x 2, y 2), C(x 3, y 3) and D(x 4, y 4). Also let x-intercept of the circle, S, be X L.vuqPNsn(Q. 39 to 40)

o`ÙkS dk lehdj.k x2 + y 2 + 2gx + 2fy + c = 0 gSA ;g o`Ùk ijoy;y2 = 4ax dks A(x 1, y 1), B(x 2, y 2), C(x 3, y 3) o D(x 4, y 4) ij feyrk gSA ekukXL o`Ùk }kjkx-v{k ij cuk;k x;k vUr%[k.M gSA

39. Identify the correct identity (identities) lgh loZ lfedk,sa gSµ (A*) y1 + y 2 + y 3 + y 4 = 0 (B*) x 1 + x 2 + x 3 + x 4 = –(8a + 4g)

(C) y 1y2y3y4 = a2

c (D*) y 1y2y3y4 = 16a2

c

40. If A, B, C are co-normal points and X L = 2 2 29 f c , then(A*) x4 = 0 (B*) x 1x2x3 = 0(C*) Circle, touches parabola (D*) (y 1 + y 2)(y2 + y 3)(y3 + y 1) = 0

;fn A, B, C lg vfHkyEc fcUnq gS rFkkXL = 2 2 29 f c , rks (A*) x4 = 0 (B*) x 1x2x3 = 0(C*) o`Ùk] ijoy; dks Li'kZ djrk gSA (D*) (y 1 + y 2)(y2 + y 3)(y3 + y 1) = 0

Sol. (39 to 40)Considering a point (at 2, 2at) and substitute it in equation of circle, S, we get

a2

t4

+ 2a(2a + g)t2

+ 4aft + c = 0

t1t2t3t4

t1 + t 2 + t 3 + t 4 = 0

t1t2 = 2

2a 2a g

a

yi = 2a ti = 0x i = a ti

2 = a{( ti)2 – 2 t1t2} = –4(2a + g)

t =2

c

a i

4

y

16a =

2

c

a yi = 16a

2c

If A, B, C are co-normal point t 1 + t 2 + t 3 = 0 and as X L = 2 (radius of S), centre lies on x-axis f = 0 and t 4 = 0 (as ti = 0)

Hence t = 0 is a repeated root of circle and parabola one of A, B, C is origin apart from D being origin, i.e. O coincides with one of the points

amongst A,B,C one of t 1, t 2, t 3 is zero t1 + t 2 = 0 or t 2 + t 3 = 0 or t 3 + t 1 = 0

and circle has double contact with parabola at origin.Sol. (39 to 40)

ekuk fcUnq(at 2, 2at) gS bls o`ÙkS ij j[kus a ij

a 2 t4 + 2a(2a + g)t 2 + 4aft + c = 0

t1t2t3t4

t1 + t 2 + t 3 + t 4 = 0


27/27



ll | |PAGE NO.-27

t1t2 = 2

2a 2a g

a

yi = 2a ti = 0x i = a ti

2 = a{( ti)2 – 2 t1t2} = –4(2a + g)

t =2

c

a i

4

y

16a =

2

c

a yi = 16a

2c

;fn A, B, C lgvfHkyEc fcUnq gS] rkst1 + t 2 + t 3 = 0 rFkk D;ksa fd XL = 2 (S dh f=kT;k), dsUæx-v{k ij gksxkA f = 0 vkS j t4 = 0 ( D;ksafdti = 0)

vr%t = 0 o`Ùk vkS j ijoy; dk iqujko ̀fÙk ewy gSA A, B, C esa ls ,d ewy fcUnq gS D ds vykokD Hkh ewy fcUnq gS] vFkkZ r~ O, A,B,C esa ls fdlh ,d lkFk

laikrh gSA t1, t 2, t 3 esa ls dksbZ ,d 'kwU; gSA t1 + t 2 = 0 ;k t2 + t 3 = 0 ;k t3 + t 1 = 0

rFkk o`Ùk vkS j ijoy; ewy fcUnq ij f}Li'khZ gSA

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