DPP 12

MATHEMATICS

Daily Practice Problems

Target IIT JEE 2010

CLASS : XIII (VXYZ) MISCELLANEOUS DPP. NO.- 12

Q.1_29/hypIf P(x₁, y₁), Q(x₂, y₂), R(x₃, y₃) & S(x₄, y₄) are 4 concyclic points on the rectangular hyperbola x y= c2_{, theco-ordinates oftheorthocentre ofthetrianglePQRare:}

(A) (x₄,!y₄) (B) (x₄, y₄) (C*) (!x₄,!y₄) (D) (!x₄, y₄)

[Hint: Arectangularhyperbolacircumscribinga5alsopassesthroughitsorthocentre

if '' ( ) * * + , i i,_tc

ct _{wherei=1, 2,3 arethevertices ofthe}5thenthereforeorthocentreis _''

( ) * * + , ! ! 3 2 1 3 2 1t t , ct t t t c ,

where t₁t₂t₃t₄= 1. Hence orthocentre is '' ( ) * * + , _! ! 4 4, _tc ct _{= (}– x₄,– y₄) ]

Q.2_4/hypLetF₁,F₂arethefociofthehyperbola ₁₆x2 ! y₉2 =1andF₃,F₄arethefoci ofitsconjugatehyperbola.

Ife_Hande_Caretheireccentricitiesrespectivelythenthestatementwhichholdstrueis (A)Theirequationsoftheasymptotesaredifferent. (B) e_H> e_C (C*)Areaofthequadrilateralformedbytheirfociis50sq. units. (D)Theirauxillarycircleswillhavethesameequation. [Hint: e_H= 5/4; e_C= 5/3 [12th_{& 13}th_11-3-2007] area = 2d1d2 = 2100 = 50 A_C: x2_{+ y}2_{= 16; A} H= x2+ y2= 9 ]

Q.3_31/hypThechordPQoftherectangularhyperbola xy=a2_{meetstheaxisof x atA; CisthemidpointofPQ}

& 'O'is theorigin.Then the5ACOis :

(A)equilateral (B*)isosceles (C)rightangled (D)rightisosceles.

[Sol. Chordwithagivenmiddlepoint 2

ky h x_{" #}

obv. OCAis isosceles with OC = CA.] Q.4_7/hypTheasymptoteofthehyperbola x_a22 y_b

2 2

! =1 formwith anytangenttothehyperbolaatrianglewhose

areaisa2_{tanBinmagnitudethenitseccentricityis:}

(A*) secB (B) cosecB (C) sec2B (D) cosec2B

[Hint: A = ab = a2_{tanB $ b/a = tanB, hence e}2_{= 1 + (b}2_/a2_{) $ e}2_{= 1 + tan}2_{B8$8e = secB8]}

Q.5_34/hypLatusrectumoftheconicsatisfyingthedifferentialequation, xdy+ydx =0andpassingthroughthe point (2,8)is :

passes through (2,8) $ c = 16

xy=16 LR = 2a(e2_{– 1) = 2a (e = 2 )}

solvingwith y=x vertex is(4, 4)

distancefrom centreto vertex =_{4 2} L.R. = length ofTA=_{8 2} Ans ]

Q.6_41/hypABisadoubleordinateofthehyperbola 1 by a x 2 2 2 2 # ! suchthat5AOB(where'O'istheorigin)isan equilateraltriangle,thentheeccentricityeofthehyperbolasatisfies (A) e > ₃ (B) 1 < e < 2_{3 (C) e = 3}2 (D*) e > 2₃ [Sol. _ax₂2 !_by₂2 #1 where y =_l 2 2 2 2 b 1 a x _{# "} l _{88$ x2= (b2+l2)} 2 2 b a _....(1) now x2_+l2_{= 4l}2 _{$ x}2_{= 3l}2 _....(2) from (1) and (2) 2 2₂ 2 3 2 b ) b ( a "l _{# l} $ a2b2+ a2l2= 3b2l2 l2(3b2– a2) = a2 b2 l2= 0 a b 3a2b 2 2 2 O ! $8 3b 2_{– a}2_{> 0 $8} 3 1 a b 2 2 O ; 1 + 3 4 a b 2 2 O $ e2 > ₃4 $88e > 2₃ Note: _abO 1_{3 $} 3 4 a b 1" ₂2 O $ e2 O ₃4 $ eO 2₃]

Q.7_47/hypThetangentto thehyperbola xy=c2 _{atthepoint Pintersectsthex-axisat Tandthey-axisatTH.The}

normal to thehyperbola at P intersects thex-axis at N and they-axis at NH. Theareas ofthe triangles

PNT and PN'T' are58and85' respectively, then 1 1_' 5 "

5 is

(A) equal to 1 (B) depends on t (C*) depends on c (D) equal to 2

[Sol. Tangent: _ctx "yt_c #2 put y = 0; x = 2ct (T) x = 0; y = t2 (T')c |||ly normal is y– t c =t2_{(x– ct)} put y = 0; x = ct– ₃ tc (N) x = 0; tc – ct3 (N')

Area of 58PNT = ' ( ) * + , " ₃ tc ct t2 c $ 5= 2 ₄ 4 t2 t ) 1( c " area of 5PN'T' = ' ( ) * + , "_ct3 t c ct $ 5' = 2 t ) 1( c2 _" 4 % 1 1_' 5 " 5 = c 1( t ) 2 ) t 1( c2 t2 4 2 4 4 " " " = c 1( t ) 2 4 2 _" (t4+ 1) = _c2₂ whichisindependentof t.]

Q.8_50hypAt the point of intersection of therectangularhyperbola xy= c2_{andthe parabola y}2_{=4ax tangents}

to therectangularhyperbolaand theparabola makean angle6and? respectivelywith theaxisof X,

then

(A*)6= tan–1(_{– 2 tan?8}) (B) _?= tan–1(_{– 2 tan68})

(C) 6= 21 tan–1(_{– tan?8}) (D)_?

= 21 tan–1(_{– tan68}) [Sol. Let (x₁, y₁) be the point of intersection $ y₁2#4ax₁ and x₁y₁= c2

y2_{= 4ax xy = c}2 % _dxdy#2_ya _dxdy#! _xy 1 ) y, x ( y a 2 tan dx dy 1 1 # ? # 1 1 ) y, x ( x y tan dx dy 1 1 ! # 6 # % _tantan ₂y_a//_yx _2axy ₂4ax_ax 2 1 1 1 2 1 1 1 1 _# ! _{#! # !} ! # ? 6 $ 6= tan–1(_{– 2 tan?8}) ]

Q.9_19/hypLocus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c2 _is

(A*) y + mx = 0 (B) y!mx = 0 (C) my!x = 0 (D) my + x = 0

[Hint: equation of chord with mid point (h, k) is _hx"_ky = 2 ; m =–

h

k _{$ y+ mx = 0 ]}

Q.10_20/hypThe locus of the foot of the perpendicular from the centre of the hyperbola xy= c2_{on a variable}

tangentis:

(A) (x2_!y2₎2_{= 4c}2_{xy (B) (x}2_{+ y}2₎2_{= 2c}2_xy

(C) (x2_{+ y}2_{) = 4x}2_{xy (D*) (x}2_{+ y}2₎2_{= 4c}2_xy

[Hint: hx + ky = h2_{+ k}2_{. Solve it with xy = c}2_{& D = 0}

Q.11_25/hypTheequation to thechord joiningtwopoints(x₁,y₁)and (x₂,y₂)ontherectangularhyperbola xy= c2_{is :} (A*) _{x x}x 1" 2 + y y y1" 2 = 1 (B) x x x1! 2 + y y y1! 2 = 1 (C) _{y y}x 1" 2 + y x x1" 2 = 1 (D) x y y1! 2 + y x x1! 2 = 1

[Hint: notethat chord of xy= c2 _{whosemiddlepoint is (h, k)in} 2

ky hx" # further, now 2h = x₁+ x₂and 2k = y₁+ y₂ ]

Q.12 Atangenttotheellipse x₉2 "y₄2 #1meetsitsdirectorcircleatPandQ.Thentheproductoftheslopes of CP and CQ where'C' is the origin is

(A) 49 (B*) 9!4 (C) 92 (D)–

4 1

[Sol. Theequation ofthetangent at (3 cos6, 2 sin6)on 1

4 y 9 x2 2 # " is 1 sin 2 y cos 3 x _{6" 6# ...(i)} Theequationofthedirectorcircleis x2_{+ y}2_{= 9 + 3 = 13 ...(ii)} ThecombinedequationofCPandCQisobtainedbyhomogenisingequation(ii)with(i).Thuscombined equationis x2_{+ y}2_{= 13} sin 2 2 y cos 3 x _' ( ) * + , _{6" 6}

$ 13₉ cos2 1'x2"13₃ sin6cos6xy ( ) * + , _6! + 13₄ sin2 1'y2#0 ( ) * + , _6!

%Product of the slopes of CP and CQ

1 sin 4 13 1 cos 9 13 y of t coefficientofx coefficien 2 2 2 2 ! 6 ! 6 # ₌ 94 4 sin 13cos 9 13 2 2 & ! 6 ! 6 = ₉13₁₃cos_cos2 9_{4 9}4 ₉4 2 ! # & ! 6 ! ! 6 ]

Q.13 The foci of a hyperbola coincide with the foci of the ellipse ₂₅x2 "y₉2 #1. Then the equation of the hyperbolawitheccentricity2is (A) ₁₂x2 !y₄2 #1 (B*) 1 12y 4 x2 2 # ! (C) 3x2– y2+ 12 = 0 (D) 9x2– 25y2– 225 = 0

[Sol. For the ellipse, a2_{= 25, b}2_{= 9}

% One of the foci is (ae, 0) i.e. (4, 0) % Forthehyperbola a'e' = 4 $ 2a' = 4 $ a' = 2 and b'2_{= 4(e'}2_{– 1) = 4 × 3 = 12} % equation ofthehyperbolais 1 12y 4 x2 2 # ! Ans.]

Paragraph for question nos. 14 to 16

Fromapoint'P'threenormalsaredrawntotheparabolay2_{=4xsuchthattwoofthemmakeangleswith}

the abscissa axis, the product of whose tangents is 2. Supposethe locus of the point 'P' is a part of a conic'C'.NowacircleS =0is describedonthechordoftheconic'C'asdiameterpassingthroughthe point(1, 0) and with gradient unity. Suppose (a, b) are the coordinates of the centre of this circle. If L₁

andL₂arethetwoasymptotesofthehyperbolawithlengthofitstransverseaxis2aandconjugateaxis 2b(principalaxesofthehyperbolaalongthecoordinateaxes)thenanswerthefollowingquestions. Q.14_404/hypLocus of P is a

(A)circle (B*)parabola (C)ellipse (D)hyperbola

Q.15 Radius ofthecircleS =0 is

(A*) 4 (B) 5 _{(C) 17 (D) 23}

Q.16 Theangle7 -(0,</2)betweenthetwo asymptotesofthehyperbolalies intheinterval

(A) (0, 15°) (B) (30°, 45°) (C) (45°, 60°) (D*) (60°, 75°)

[Sol. Equationofanormal y=mx– 2m – m3 [12th& 13th03-03-2007]

passesthrough(h, k)] m3_{+ (2– h)m + k = 0} m₁m₂m₃=– k but m₁m₂= 2 $ m₃=– k/2 thismustsatisfyequation(1) 8 k3 – (2 – h) 2 k +k=0 k3_{– 4k(2 – h) + 8k = 0 (k =0)} k2_{– 8 – 4h + 8 = 0}

locus of 'P' is y2_{=4x which is a parabola} Ans.

now chordpassingthrough(1,0)isthefocalchord. Giventhatgradientoffocalchordis1 % 2 1 t t "2 = 1 $ t1+ t2= 2, Also t1t2=– 1

equationofcircledescribed ont₁t₂asdiameteris (x– t )(x₁2 – t ) + (y2₂ – 2t₁)(y– 2t₂) = 0 x2_{+ y}2_{– x(} 2 1 t + 2 2 t ) + 2 1 t 2 2 t – 2y(t₁+ t₂) + 4t₁t₂= 0 x2_{+ y}2_{– x[4 + 2] + 1 – 2y(2) – 4 = 0} x2_{+ y}2_{– 6x – 4y – 3 = 0}

now the hyperbola is 9x2 –

4 y2

= 1 asymptotes are y = 32 andy=x –

3x 2 now tan6= 2/3

% 7= 26

tan7= ₁2_!·(₍2₄ )3₎₉; tan7= 512 ; 7= tan–1 '

( ) * + , 5 12 hence 7 -(60°, 75°) Ans. ] Paragraph forquestionnos.17 to 19 AconicCpassesthroughthepoint(2,4)andissuchthatthesegmentofanyofitstangentsatanypoint contained between the co-ordinate axes is bisected at the point of tangency. Let S denotes circle described on the foci F₁and F₂of theconic C as diameter.

Q.17 Vertex oftheconicC is (A) (2, 2), (–2, – 2) (B*)

3

2 ,22 2

4

,

3

!2 ,2!2 2

4

(C) (4, 4), (–4, – 4) (D)

3

,2 2

4

,

3

! ,2! 2

4

Q.18 Directorcircleoftheconicis (A) x2_{+ y}2_{= 4 (B) x}2_{+ y}2_{= 8 (C) x}2_{+ y}2_{= 2 (D*) None} Q.19 EquationofthecircleSis (A) x2_{+ y}2_{= 16 (B) x}2_{+ y}2_{= 8 (C*) x}2_{+ y}2_{= 32 (D) x}2_{+ y}2_{= 4} [Sol. Y!y = m (X!x) ; if Y= 0 then X = x! y m and if X = 0 then Y= y!mx. Hence x! y m = 2 x $ dydx =! yx (0, y–mx) (x– ,0)y_m O P(x,y) dy y " C dxx C = c $ xy = c passes through(2, 4)

$ equation ofconicis xy= 8

whichisarectangularhyperbolawithe= 2. (4,4) (–4, –4) 2 2 ,2 2 2 2 ,2 2 ! !

Hence the two vertices are

3

2 ,22 2

4

,

3

!2 ,2!2 2

4

focii are (4, 4) & (!4, 4)

Assertion and Reason

Q.20 Statement-1: Diagonalsofanyparallelograminscribedinanellipsealwaysintersectatthecentreof theellipse.

Statement-2: Centre ofthe ellipse is the onlypoint at which two chords can bisect each other and everychordpassingthroughthecentreoftheellipsegets bisectedatthecentre. (A*)Statement-1isTrue, Statement-2 isTrue;Statement-2is acorrectexplanationforStatement-1 (B)Statement-1isTrue,Statement-2isTrue;Statement-2isNOTacorrectexplanationforStatement-1 (C)Statement -1 isTrue, Statement-2 is False

(D) Statement-1 is False, Statement-2 isTrue

[Sol. Statement-2is correctasellipseisacentralconicanditalsoexplainsStatement-1. Hence, code(A)is thecorrect answer.]

Q.21 Statement-1: ThepointsofintersectionofthetangentsatthreedistinctpointsA,B,Contheparabola y2_{=4x can be collinear.}

Statement-2: IfalineLdoesnot intersect theparabolay2_{=4x,thenfromeverypointofthelinetwo}

tangents can bedrawn to the parabola.

(A)Statement-1isTrue, Statement-2 isTrue; Statement-2 isacorrectexplanationforStatement-1 (B)Statement-1isTrue,Statement-2isTrue;Statement-2isNOTacorrectexplanationforStatement-1 (C)Statement -1 isTrue, Statement-2 is False

(D*)Statement -1 is False, Statement-2 isTrue

[Sol. Areaofthetrianglemadebytheintersectionpointsoftangents at pointA(t₁),B(t₂)andC(t₃)is 0 t t t t t t 2 1 1 3 3 2 2 1! ! ! =

Hence, Statement-1 iswrong.Statement-2 is correct.

Hence, code (D)is the correct answer. ]

Q.22 Statement-1: Thelatusrectumis theshortestfocalchordinaparabolaoflength4a.

because Statement-2: Asthelengthofafocalchordoftheparabola 2 2 t 1 t a is ax 4 y ' ( ) * + , " # ,whichisminimum when t = 1.

(A*)Statement-1isTrue, Statement-2 isTrue;Statement-2is acorrectexplanationforStatement-1 (B)Statement-1isTrue,Statement-2isTrue;Statement-2isNOTacorrectexplanationforStatement-1 (C)Statement -1 isTrue, Statement-2 is False

(D) Statement-1 is False, Statement-2 isTrue

[Sol. Let P(at2_{, 2at) be the end of a focal chord PQ of the parabola y}2_{= 4ax. Thus, the coordinate of the}

otherend point Q is ' ( ) * + , _! ta 2 , ta2 % 2 ₂ 2 2 ta 2 at 2 ta at PQ ' ( ) * + , _" " ' ( ) * + , _! # 2 ₂ 2 2 t 1 t 4 t1 t ' ( ) * + , " " ' ( ) * + , ! # 4 t 1 t t 1 t a '2" ( ) * + , ! ' ( ) * + , " # 2 4 t1 t t 1 t a _' 2_{" 2 ! "} ( ) * + , " # 2 t 1 t a ' ( ) * + , " #

% Length offocal chord is, 2 t 1 t a ' ( ) * + , " , where t 1_t'22 ( ) * + , " _{forall t}=0. % a t 1_t 4a 2 2 ' ( ) * + , " _{$ PQ24a} Thus,thelengthofthefocalchordoftheparabolais4awhichisthelengthofitslatusrectum. Hence,thelatusrectumofaparabolais theshortestfocalchord.

Thus, Statement-1 andStatement-2is trueandStatement-2s correct explanationofStatement-1 ] Q.23 Statement-1: If P(2a, 0) be any point on the axis of parabola, then the chord QPR, satisfy

2 2 2 _(PR1_{) 4a}1 ) PQ ( 1 " # .

Statement-2: ThereexistsapointPontheaxisoftheparabolay2_{=4ax(otherthanvertex),suchthat} 2

2 _(PR1₎

) PQ

( 1 " = constantfor all chord QPR of the parabola.

(A*)Statement-1isTrue, Statement-2 isTrue;Statement-2is acorrectexplanationforStatement-1 (B)Statement-1isTrue,Statement-2isTrue;Statement-2isNOTacorrectexplanationforStatement-1 (C)Statement -1 isTrue, Statement-2 is False

(D) Statement-1 is False, Statement-2 isTrue

[Sol. Let P(h, 0) (where h=0)beapointon theaxis ofparabolay2=4ax thestraight linepassingthrough

Pcuts the parabola at a distance r.

$(r sin6)2= 4a (h + r cos6)

$r2sin26 – (4a cos 6)r– 4ah = 0 ...(i)

where, r₁+ r₂= 4_sinacos2₆6 and r1r2 =– _sin4 .ah2₆

% ₂ 2 2 1 2 2 _PR1 _r1 _r1 PQ1 " # " rr rr cosh sin2ah 2 2 2 2 2 2 1 2 2 2 1 " _# 6 _" 6 #

which is constant only, ifh2_{=2ah i.e., h=2a}

$ _{2 2} 2 2 2 2 2 _PR1 cos_4a sin_{4a 4a}1 PQ1 # 6 " 6 # "

Thus, _PQ12 " _PR12 =constantforall chords QPR,

if h = 2a.

Hence, (2a, 0)is therequired point on theaxisof parabola.

% Statement-1and Statement-2 aretrueand Statement-2 is correct explanationofStatement-1 ]

Q.24 Statement-1: The quadrilateral formed by the pair of tangents drawn from the point (0, 2) to the parabola y2_{– 2y + 4x + 5 = 0 and the normals at the point of contact of tangents in a}

square.

Statement-2: Theanglebetweentangentsdrawnfromthegivenpointtotheparabolais90°.

(A)Statement-1isTrue, Statement-2 isTrue; Statement-2 isacorrectexplanationforStatement-1 (B)Statement-1isTrue,Statement-2isTrue;Statement-2isNOTacorrectexplanationforStatement-1 (C)Statement -1 isTrue, Statement-2 is False

[Sol. (y– 1)2=– 4(x + 1)

Directrix x + 1 = 1

x = 0 (0,2)

x=0

If tangents are drawn from (0, 2) to the parabola(i.e. from directrix) thenlengthoftangentwillbeunequalhencethequadrilateralformedby pairoftangentsand normalsatthepointofcontactisrectangle.]

More than one are correct:

Q.25_502hypIf the circle x2_{+ y}2_{= a}2_{intersects the hyperbola xy = c}2 _{in four points P(x}

1, y1), Q(x2, y2),

R(x₃, y₃), S(x₄, y₄), then

(A*) x₁+ x₂+ x₃+ x₄= 0 (B*) y₁+ y₂+ y₃+ y₄= 0

(C*) x₁x₂x₃x₄= c4 _{(D*) y}

1y2y3y4= c4

[Sol. solving xy = c2 _{and x}2_{+ y}2_{= a}2

x2₊ 2 4 xc = a2 x4_{– ax}3_{– a}2_x2_{+ ax + c}4_{= 0} $

P

x_i#0 ;

P

y_i#0 x₁x₂x₃x₄= c4 _{$ y} 1y2y3y4= c4]

Q.26_503hypThetangenttothehyperbola, x2_!3y2_{=3 atthepoint}

3

_{3 0,}

4

_{whenassociatedwithtwoasymptotes}

constitutes:

(A) isoscelestriangle (B*) anequilateraltriangle

(C*) atriangleswhoseareais 3 sq. units (D) arightisoscelestriangle.

[Hint: area of the5= ab sq units ; H : x2/3– y2/ 1 = 1 ]

Q.27 The locus of the point of intersection of those normals to the parabola x2=8y whichare at right

angles to each other,is aparabola.Whichofthefollowinghold(s)goodin respectofthelocus? (A*)Lengthofthelatusrectumis2.

(B) Coordinates offocus are '

( ) * + , 2 11 ,0

(C*) Equation of a directro circleis 2y– 11 = 0

(D)Equationofaxis ofsymmetryy=0.

Match the column:

Q.28₁₀₅ Column-I Column-II

(A) Ifthechord ofcontact oftangents fromapointPto the (P) Straightline

parabola y2_{= 4ax touches the parabola x}2 _{= 4by, the locus of P is}

(B) AvariablecircleChastheequation (Q) Circle

x2_{+ y}2_{– 2(t}2_{– 3t + 1)x – 2(t}2_{+ 2t)y + t = 0, where t is a parameter.}

Thelocusofthecentreofthecircleis (C) Thelocusofpointofintersectionoftangentstoanellipsex_a22 y_b 2 2 " =1 (R) Parabola attwopointsthesumofwhoseeccentricanglesisconstantis (D) Anellipseslidesbetweentwoperpendicularstraightlines. (S) Hyperbola Thenthelocusofitscentreis [Ans. (A) S; (B) R; (C) P; (D) Q] [Sol.

(A) yy₁= 2a (x +x₁) ; x2_{= 4by = 4b[(2a/y}

1)(x + x1)] $y1x2!8abx!8abx1= 0 ;

D = 0 gives xy =!2ab $ Hyperbola

(B) centre is x = t2_{– 3t + 1 ....(1) [18-12-2005, 12}th_] y = t2_{+ 2t ....(2)} (2)– (1) gives – x + y = 5t – 1 or t = 1!x₅"y Substitutingthevalueof tin(2) y = y ₅x 1'2 ( ) * + , ! " + 2 ' ( ) * + , ! " 5x 1 y 25y = (y– x + 1)2+ 10(y– x + 1)

25y = y2_{+ x}2_{+ 1– 2xy – 2x + 2y + 10y – 10x + 10}

x2_{+ y}2_{– 2xy – 12x – 13y + 11 = 0} whichisaparabola as5 =0 and h2= ab ] (C) h = 2 cos 2 cos a > ! 7 > " 7 ; k = 2 cos 2 sin b > ! 7 > " 7 given 27"> =constant=C 2 cos7!> % = k C sin b h C cos a _{# $ y = tanC x} a b _' ( ) * + , Locusof(h,k)isastraightline (D) y₁y₂= x₁x₂= b2 _....(1) and (x₂– x₁)2+ (y₂– y₁)2= 4(a2– b2) ....(2) Also 2h = x₁+ x₂ 2k = y₁+ y₂ from (2) (x₁+ x₂)2_{+ (y} 1+y2)2– 4(x1x2+ y1y2) = 4(a2– b2) 4@ (h2+ k2)– 4@ (2b2) = 4@ (a2– b2) % x2+ y2= a2+b2$ Circle

Alternative: Equationofdirectorcirclewithcentre(h,k) (x– h)2+ (y– k)2= a2+ b2

(0, 0)lies on it $ h2+ k2= a2+ b2 $ locus is x2+ y2= a2+ b2 ]

Q.29₁₀₆ Column-I Column-II

(A) Foran ellipse x₉2 " y₄2 #1withverticesAandA', tangentdrawnatthe (P) 2 point Pin thefirst quadrantmeetsthey-axis inQandthechordA'Pmeets

the y-axis in M. If'O' is the origin then OQ2_{– MQ}2_{equals to}

(B) Iftheproductoftheperpendiculardistancesfromanypointonthe (Q) 3

hyperbola x_a22 ! _by₂2 #1ofeccentricitye= 3 fromitsasymptotes isequalto6,then thelengthofthetransverseaxisofthehyperbolais

(C) Thelocusofthepointofintersectionofthelines (R) 4

3x!y!4 3 t = 0 and 3tx + ty!4 3 = 0

(wheretis aparameter)isahyperbola whoseeccentricityis

(D) If F₁& F₂are the feet ofthe perpendiculars from the foci S₁& S₂ (S) 6 ofanellipsex₅2 "y₃2 =1onthetangentatanypoint Pontheellipse,

then (S₁F₁). (S₂F₂) is equal to [Ans. (A) R; (B) S; (C) P; (D) Q]

[Sol.(A) a = 3 ; b = 2

T : xcos₃ 6"ysin₂ 6 #1 x = 0 ; y = 2 cosec6

chordA'P, y = _3(cos2sin ₎₁(x" )3 "

6 6

put x = 0 y =₁2_"sin_cos6₆ = OM

Now OQ2_{– MQ}2_{= OQ}2_{– (OQ – OM)}2_{= 2(OQ)(OM)– OM}2_{= OM{ 2(OQ)– (OM) }}

= ₁2_"sin_cos6_{6 DE}F ! _" 6_69:; 6 1 cos sin 2 sin4 = 9: ; D E F 6 " 6 6 ! ! 6 " 6 " 6 ) cos 1(

sincos ) 1( cos ) 1(

2 cos 14sin

2

= 41(₁₍"_"cos_cos6₆)(₎₍2₁!_"1_cos"cos₆₎6) = 4

(B) p₁p₂= _aa₂2_"b_b2₂ = 2 2 2 2 2 e a e( )1 a. a ! = 6; 6 3 a 2 2 # $ a2= 9 $ a = 3 hence 2a = 6 (C) hyperbola ₁₆x2 !₄₈y2 #1 (D) Productofthefeetoftheperpendicularsis equaltothesquareofitssemiminoraxes.]