MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. TIME : 3hrs M. Mrks.75 Note: This question pper consists of three sections A,B nd C. SECTION A VERY SHORT ANSWER TYPE QUESTIONS. X = ) Find the eqution of circle with centre (, 3) nd touching the line 3-4 + =.. Find k if the following pir of circles re orthogonl. + + -k =, + ++8= 3. A doule ordinte of the curve = 4 is of length 8. Prove tht the line from the verte to its ends re t right ngles. 4. Find the condition for the line cos + sin = p to e tngent to the ellipse. 5. If the ngle etween smptotes is 3 then find its eccentricit. 6. Evlute sec d 7. Evlute sin d sin( ) on I R {n : n Z}. 8. Evlute 7 / ( ) d
9. Evlute n lim... n n 6n. Find the generl solution of d d. SECTION B SHORT ANSWER TYPE QUESTIONS. ANSWER ANY FIVE OF THE FOLLOWING 5 X 4 =. Find the eqution of tngents the circle + = t the points whose sciss re. Find the eqution of the circle which psses through the point (, -3) nd intersects the circles given the equtions + 6+3+5 =, + + 7 = orthogonll. 3) If the Stright line + 3 = intersects the circle + = 4 t the points A nd B. Find the eqution of the circle hving AB s dimeter. 4. Show tht the point of intersection of perpendiculr tngents to n ellipse lie on circle. 5. Find the eqution of the hperol of given length of trnsverse is is 6 whose verte isects the distnce etween the centre nd the focus. 6. Evlute /4 log( tn )d 7.find the differentil eqution corresponding to the fmil of curves eliminting ritrr constnts given the eqution + = ; (, )
SECTION C LONG ANSWER TYPE QUESTIONS. ANSWER ANY FIVE OF THE FOLLOWING 5 X 7= 35 8. Find the locus of the point whose polrs with respect to the circles + 4-4 - 8 = nd + + 6 = re mutull perpendiculr. 9. Find the eqution of circle pssing through the points (3, 4); (3,); (,4). Prove tht the re of the tringle inscried in the prol = 4 is ( )( 3 )( 3 ) 8 vertices. sq.units where,, 3 re the ordintes of its. Evlute 4 d on R.. Evlute d sin cos 3. Show tht the re of the region ounded deduce the re of the circle. 4. solve ( )d = ( 3 3 )d (ellipse ) is. Also
SOLUTIONS. Find the eqution of circle with centre (, 3) nd touching the line 3-4 + =. Sol. Centre C=(,3). Rdius r =Perpendiculr distnce from C to 3-4 += = Eqution of circle (-h) + (-k) = r (-) + (-3) = + 4 6 + = P 3 4 + =. Find k if the following pir of circles re orthogonl. + + -k =, + ++8= Sol. Given circles re + + -k =, + ++8= from ove equtions g = ; f =; c g = ; f = ; c = 8 since the circles re orthogonl, k g g + f f = c + c () () + () () k + 8 = - k + 8 3. A doule ordinte of the curve = 4 is of length 8. Prove tht the line from the verte to its ends re t right ngles. Sol. Let P = (t, t) nd P = (t, t) e the ends of doule ordinte PP. Then 8 PP (4t) 4t t P = (4, 4), P = (4, 4) Slope of AP slope of AP = 4 4 4 4
PAP 4. Find the condition for the line cos + sin = p to e tngent to the ellipse. Sol. Eqution of the ellipse is Eqution of the line is cos + sin = p sin = cos + p cos p sin sin cos m,c sin p sin Aove line is tngent to the ellipse c = m + p cos (i) sin sin or p = cos + sin. 5. If the ngle etween smptotes is 3 then find its eccentricit. Sol: Angle etween smptotes of hperol is sec e. sec e 3 sec e 5 e sec5 cos5 cos(45 3 ) cos45 cos3 sin 45 sin3 3 ( 3 ) 3 ( 3 )( 3 ) ( 3 )
6. 7. ( 3 ) 6. sec d sec (sec tn ) d sec tn sec sec tn d log sec tn c sec tn sin sin log c log c cos cos cos Sol: 8. cos( / ) log sin( / ) log sin log tn c sin 4 cos 4 4 4 c sin d on I R {n : n Z}. sin( ) sin sin( ) d d sin( ) sin( ) sin( )cos cos( )sin d sin( ) cos( ) cos d sin d sin( ) cos sin log sin( ) c. 7 / ( ) d Sol. = sin = sin d = cos d = / / 7 / sin ( sin ) cos d / / 9 8 9 8 cos sin d cos sin d c
9. Sol: 9 / 9 9 cos 9 9 9 9 lim... n n n 6n lim... n n n n 5n lim... n n 5n n n n 5 5 d log( ) log 6. Find the generl solution of d d. Sol. Given d.e. is d d d d Integrting oth sides d d sin = sin + c Solution is sin + sin = c, where c is constnt.. Find the eqution of tngents the circle + = t the points whose sciss re Sol. Eqution of the circle is S = + = Let the point e (, ) + = = 9 Y = 3. Co ordintes of P re (,3) nd (, -3) Eqution of the tngent t P (, 3) is S =.. +. 3 = + 3 = Eqution of the tngent of P(, -3) is S =. + (-3) = 3 =
. Find the eqution of the circle which psses through the point (, -3) nd intersects the circles given the equtions + 6+3+5 =, + + 7 = orthogonll. Sol. Let circle e + + g + f + c = -------() () is orthogonl to + 6+3+5 = g( 3) + f 3 = c + 5 6g + 3f = c + 5-----------() () is orthogonl to + 7 = g + f 7 = c g 7f = c ----(3) Given () is pssing through (, 3) + 9 6f + c = (3) () 5g f = 5 g f = (iii) + (iv) 9 g 3f = g + 3f = 9 g f = 5f = f = g =. g = + 9 6. + c = c = 5 Therefore, eq of the circles re + + + 5 = (or) 3 + 3 + + 4 5 = 3) If the Stright line + 3 = intersects the circle + = 4 t the points A nd B. Find the eqution of the circle hving AB s dimeter. Sol. circle is S = + = 4 Eqution of the line is L = + 3 = Eqution of circle pssing through S= nd L= is S+ L = ( + 4) + ( + 3 ) =
+ + + 3 4- = Center Centre lies on + 3 = (- ) + 3 - = = Eqution of circle e 3 ( + )-4 3 ( + 3 -) = 3 ( + ) - 4 6-5 =. 4. Show tht the point of intersection of perpendiculr tngents to n ellipse lie on circle. Sol. Eqution of the ellipse is Let P(, ) e the point of intersection of the tngents. Eqution of the tngent is m m This tngent is pssing through P(, ) m m m m ( m ) m m m m m ( ) m ( ) This is qudrtic eqution in m giving two vlues for m s m nd m. These re the slopes of the tngents pssing through (, ). The tngents re perpendiculr m m = Locus of P(, ) is + = + which is circle. Q P(, ) R
This circle is clled Director circle of the Ellipse. 5. Find the eqution of the hperol of given length of trnsverse is is 6 whose verte isects the distnce etween the centre nd the focus. Sol. M P S A C Z A N S L= Let the hperol e Given CA = AS 6. Sol. = e = e e = Length of trnsverse is is = 6 = 3 = (e ) = 9(4 ) = 7 Eqution of the hperol is 9 7 /4 log( /4 3 7 tn )d I log tn d 4 /4 /4 log log tn 4 tn tn tn 4 tn tn d d
/4 /4 log tn tn tn log log( tn )d / 4 / 4 d log d log( tn )d /4 log () I log I log 4 8 7. + = ; (, ) Sol. Given eqution is + = ------() Differentiting w.r.t. I Differentiting w.r.t. ----(3) 3 8. Find the locus of the point whose polrs with respect to the circles + 4-4 - 8 = nd + + 6 = re mutull perpendiculr. Sol. Eqution of the circles is S= + 4-4 - 8 = - () S = + + 6 = - () let P (, ) e n position in the locus. Eqution of the polr of p w.r.to circle () is ( + ) ( + ) 8 = ( -) + ( -) ( + + 8) = (3) Polr of P w.r. to circle () is + (+ ) 3 ( + Y ) =
+ - + 3 + 3 = ( ) + ( + 3) ( + 3 + ) = (3) nd (4) re perpendiculr + = ( ) ( ) + ( ) ( +3) = 3 6 Locus of p(, ) is + - 3 + 4 = 9. Find the eqution of circle pssing through the points (3, 4); (3,); (,4) Let the eqution of circle e + + g + f + c = it is pssing through (3, 4); (3,); (,4) Given points stisf ove eqution then 9 + 6 + 6g + 8f + c = 5+6g + 8f + c = (i) 9 + 4 + 6g + 4f + c = 3 + 6g + 4f + c = (ii) +6+g +8f +c = 7+g +8f +c = (iii) (ii) (i) we get - -4f = (or) f = - 3 (ii) (iii) we get -4 +4g -4f = g f = => g = - Now sustituting g, f in eqution (i) we get 5 + 6 (-) +8 (-3) + c = We get c = Required eqution of circle e X + -4 6 + =. Prove tht the re of the tringle inscried in the prol = 4 is ( )( 3 )( 3 ) 8 vertices. Sol. Given prol = 4 let sq.units where,, 3 re the ordintes of its P(t,t ), Q(t,t ), R(t 3,t 3) e the vertices of PQR. Are of PQR = 3 t t t t t t (t t 3 ) t t 3 (t t ) t t t t 3 (t t )(t t ) t t t t 3 3
. Sol: (t t )(t t )(t t ) 3 3 3 (t t )(t t 3 )(t 3 t ) (t t )(t t 3 )(t 3 t ) 8 ( )( 3 )( 3 ) 8 Where P(, ), Q(, ), R( 3, 3 ) re the vertices of PQR. 4 Tke d on R. d 4 d. d ( ( ) ) t then d dt dt dt d t t ( ) 4 t tn tn c c. Sol. d sin cos d sin cos tn c.
= d tn tn tn tn sec d tn tn tn sec put tn t sec d dt tn dt dt log t C t t log tn C 3. Show tht the re of the region ounded deduce the re of the circle. (ellipse ) is. Also Sol: The ellipse is smmetricl out X nd Y is Are of the ellipse = 4 Are of CAB= 4. 4 Eqution of ellipse is
CAB = dn = sin =.. 4 4 (from pro. 8 in e () ) = Sustituting =, we get the circle Are of the circle = 4. ( )d = ( 3 3 )d Sol. ( )d = ( 3 3 )d d d 3 3 Put = v so tht d sq. units. which is homogeneous d.e. d 3 3 v dv v v d 3 3v 3 dv v v d 3v v dv d 3 (v v ) v v 3 ( 3v) 3v v v v v( 3v) v 3v 3v 3v dv v 3v d dv d 3v v 3 d dv v v 3log v log log c v
3log log log c log log c 3 log log c 3 3 3 3 log c log c c 3 3 / 3 e c / 3 / c e e