MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. Tim: 3hrs Ma. Marks.75 Not: This qustion papr consists of thr sctions A, B and C. SECTION -A Vry Short Answr Typ Qustions. 0 X = 0. Find th condition that th tangnts Drawn from (0,0) to S + y + g + fy + c = 0 b prpndicular to ach othr.. Find th lngth of th chord intrcptd by th circl +y + 3y = 0 on th lin y = -3 3. Show that + y + l + 4 = 0; + y + my - g = 0 Circls intrsct ach othr orthogonally. 4. Find th vrt and focus of 6 6y+6 = 0. 5. If 3 4y + k = 0 is a tangnt to 4y = 5, find valu of k. log 6. Evaluat + d on (0, ). 7. Evaluat d, R + 8. Evaluat π/ π / + cos d 9. Evaluat π/ 4 0 sc 4 θdθ
dy y + 0. Solv = d + y 3 SECTION -B Short Answr Typ Qustions. Answr Any Fiv of th Following 5 X 4 = 0. If th two circls + y + g + fy = 0 and + y + g + f y = 0 touch ach othr thn show that f g = fg.. Find th quations of th tangnts to th circl +y + y 3 = 0 which ar prpndicular to 3 y+4=0 3. Find th condition for th lin l + my + n = 0 to b a normal to th llips a y + = b 4. Find th quation of th tangnts to 9 + 6y = 44 which maks qual intrcpts on coordinat as. 5. If, b th ccntricity of a hyprbola and its conjugat hyprbola thn + =. 6. Evaluat (n!) lim n n / n 7. Solv dy ( + ) + y = d tan
Long Answr Typ Qustions. SECTION - C Answr Any Fiv of th Following 5 X 7 = 35 8. Find th quation of th circl passing through (0,0) and making intrcpts 4,3 on X ais and Y ais rspctivly. 9. Show that th quation to th pair of tangnts to th circl S = 0 from P(, y ) is S = S S. 0. Prov that th tangnts at th trmitis of a focal chord of a parabola intrsct at right angls on th dirctri.. (6 + 5) 6 + d. Evaluat d 4 + 3. Find th ara boundd btwn th curvs y 4a, 4by( a 0,b 0) = = > >. 4. Solv (y y)d + (y )dy = 0
Maths B Papr 3 - Solutions. Find th condition that th tangnts Drawn from (0,0) to S + y + g + fy + c = 0 b prpndicular to ach othr. Sol. Lt th angl btwn th pair of Tangnts thn tan = givn π θ =, radius r= g + f c S = + y + g + fy + c =0+c =c tan 45 0 = g + f c + y + g + fy + c = g + f c = c. Find th lngth of th chord intrcptd by th circl +y + 3y = 0 on th lin y = -3 Sol. Equation of th circl is S + y + 3y = 0. Cntr C (, - ) and Radius r = g f c = = = Equation of th lin is y = 3 y 3 = 0 P = distanc from th cntr to th lin = = Lngth of th chord = = = = 96 = 4 6 units
3. Show that + y + l + 4 = 0; + y + my - g = 0 circls intrsct ach othr orthogonally. Sol. Givn circls + y + l + 4 = 0; + y + my - g = 0 from ths quations, g = - l; f = 0, c = g, g = 0, f = m, c = - g Now g g + f f = c + c (-l) (0) + (0) (m) = g g 0 = 0 Two circls ar orthogonal. 4. Find th vrt and focus of 6 6y+6 = 0. Sol. Givn parabola is 6 6y + 6 = 0 6 = 6y 6 ( 3) - 9 = 6y 6 ( 3) = 6y + 3 ( 3) = 6 y + = 6 y h = 3, k =, a = 6 = 3 4 Vrt = (h, k) = 3, Focus = (h, k+a) = 3 3, + = (3,) 5. If 3 4y + k = 0 is a tangnt to 4y = 5, find valu of k. Sol. Equation of th hyprbola 4y = 5 y 5 = a = 5,b = 5 (5 / 4) 4
Equation of th lin is 3 4y + k = 0 4y = 3 + k y = 3 + k ---- () 4 4 3 k m =, c = 4 4 If () is a tangnt to th hyprbola thn c = a m b k 9 5 = 5 6 6 4 k = 45 0 = 5 k = ± 5 log 6. Evaluat + d on (0, ). + log Sol. d = log + d = log + C 7. Evaluat d, R + Sol. d + = d + + = d = log( + ) + C + 8. Evaluat Sol. Lt I = π/ π / + cos d π/ cos d (i) π/+
π/ b b cos( π / π / )d I = f ()d = f (a + b )d π/ + a a Adding () and (), π/ cos d = () + π/ π/ π/ cos ( + ) I = d = cos d π/ 0 π/ + π/ [ ] π/ 0 ( ) I = cos d cos is vn function I = sin I = 9. Evaluat π/ 4 0 sc 4 θdθ π/4 π/4 π/4 4 sc θdθ = sc θ.sc θdθ = sc θ ( + tan θ)dθ 0 0 0 π/4 π/4 π/4 Sol. Lt ( ) = sc θ + sc θ tan θ dθ = sc θdθ + tan θsc θdθ 0 0 0 3 π/4 π/4 tan θ 4 o ( ) 3 3 3 o = tan θ ) + = 0 + 0 = 0. Solv dy y + = d + y 3 Sol. b =, a = b = a ( + y 3)dy = ( y + )d ( + y 3)dy ( y + )d = 0 (dy + y d) + y dy 3 dy d d = 0 Intgrating : y + y 3 = c
. If th two circls + y + g + fy = 0 and + y + g + f y = 0 touch ach othr thn show that f g = fg. Sol. S = + y + g + fy = 0 Cntr C = ( g, f), radius r = g + f S = + y + g + f y = 0 C = ( g, f ), r = g + f Givn circls ar touching circls, C C = r + r (C C ) = (r + r ) (g g) + (f f ) = g + f + g + f + g + f g + f / (gg + ff ) = {g g + f f + g f + f g } (gg + ff ) = g g + f f + g f + f g g g + f f + gg ff = g g + f f + g f + f g gg ff = g f + f g g f + f g gg ff = 0 (gf fg ) = 0 gf = fg. Find th quations of th tangnts to th circl +y + y 3 = 0 which ar prpndicular to 3 y+4=0 Sol. S = +y +-y-3=0, cntr C(-, ) and radius r = + + 3 = 5 Equation of th lin prpndicular to 3-y+4 =0 is + 3y +k = 0 5= + 3 + k + 9 ( k + 5 = ) 0
50 = k +4k+4 k +4k-46 =0 4 ± 6 + 84 k = 4 ± 0 k = ± 5 Equation of th rquird tangnt is + 3y 5 =0 3. Find th condition for th lin l + my + n = 0 to b a normal to th llips a y + =. b Sol. Equation of th llips is a y + = b Lt l + my + n = 0 b normal at P(a) Equation of th normal at P(a) is : a by a b cos θ sin θ = () L + my = n () Comparing () and () l = m = n a b a b cos θ sin θ lcos θ msin θ n = = a b a b an bn cos θ =, sin θ = l(a b ) m(a b ) cos θ + sin θ = a n b n + = l (a b ) m (a b ) a b (a b ) + = is th rquird condition. l m n
4. Find th quation of th tangnts to 9 + 6y = 44 which maks qual intrcpts on coordinat as. Sol. Equation of th llips is 9 + 6y = 44 y + = 6 9 Equation of th tangnt is cos θ + y sin θ = a b Slop of th tangnt = b cos θ = a sin θ a 4 cot θ = = b 3 4 3 cos θ = ±,sin θ = ± 5 5 Equation of th tangnt is: 4 y 3 ± + ± = 4 5 3 5 ± y ± 5 = 0 5. If, b th ccntricity of a hyprbola and its conjugat hyprbola thn + =. Sol. Equation of th hyprbola is S= a y = b + + = a b = a b a a = a a + b... ()
Equation of th conjugat hyprbola is y y = = a b b a a + b a + = b = b Adding () and () b b = a b +... () a b a + b + = + = = a b a b a b + + + 6. Evaluat (n!) lim n n / n (n!) Sol: lim n n / n / n ( n) = lim n n / n ( n) = lim n n n ( n) Lt y = lim n n n / n i n log y = lim log...... n n n n n n = lim n n n r r= log n r = lim log n n n
[ log ] [ ] [ ] = log = log d 0 0 0 = = (log ) = y = =. 0 0 7. Solv dy ( + ) + y = d tan Sol. tan dy + y = d + + which linar diffrntial quation in y. d pd + tan I.F. = = = Sol is y.i.f. = y.i.f = Q. I.F. d tan tan ( ) y = d () + Considr tan ( ) + d d put tan = t = dt + t tan t t = ( ) dt = dt = = Solution is tan tan c y = + tan tan y = + c 8. Find th quation of th circl passing through (0,0) and making intrcpts 4,3 on X ais and Y ais rspctivly. Sol.
Lt th quation of th circl b + y + g + fy + c = 0. Givn circl is making intrcpts 4, 3 on, y as rspctivly. Thrfor, (4,0) and (0,3) ar two points on th circl. Circl is passing through (0,0), (4,0) and (0,3). (0,0) 0 + 0 + g(0) + f(0) + c = 0 C = 0 (4,0) 6+0+8g+f.0+c =0 G = as c = 0 (0,3) 0+ 9+g.0+6f +c = 0 f = - as c = 0 Rquird quation of circl is X + y -4 3y = 0 9. Th quation to th pair of tangnts to th circl S = 0 from P(, y ) is S = S S. Proof:
Lt th tangnts from P to th circl S=0 touch th circl at A and B. Equation of AB is S =0. i.., ( ) ( ) + y y + g + + f y + y + c = =====i) 0 Lt Q(, y ) b any point on ths tangnts. Now locus of Q will b th quation of th pair of tangnts drawn from P. th lin sgmnt PQ is dividd by th lin AB in th ratio -S :S PB = QB S S ----ii) BUT PB = S, QB = S PB QB S = ----iii) S From ii) and iii) s S = s S S S = S Hnc locus of Q( y ) is S S = S, 0. Prov that th tangnts at th trmitis of a focal chord of a parabola intrsct at right angls on th dirctri. Sol. Lt th parabola b y = 4a Equation of th tangnt at P(t ) is = + t y at
Equation of th tangnt at Q(t ) is = + t y at Solving, point of intrsction is T[att,a(t + t )] Equation of th chord PQ is (t + t )y = + att Sinc PQ is a focal chord, S (a,0) is a point on PQ. Thrfor, 0 = a +a t t t t = -. Thrfor point of intrsction of th tangnts is [ a,a(t + t )]. Th coordinat of this point is a constant. And that is = -a which is th quation of th dirctri of th parabola. Hnc tangnts ar intrscting on th dirctri.. (6 + 5) 6 + d Sol. ( ) d lt 6+5=A 6 + + B d 6 + 5 = A( 4) + B Equating th cofficints 3 6 = 4A A = Equating th constants A + B = 5 B = 5 A = 3 3 5 + =
(6 + 5) 6 + d 3 3 = ( 4) 6 d 6 d + + + 3/ 3 (6 + ) 3 = + 3 d 3/ + 3/ 3 7 = (6 + ) + d 4 4 3/ 3 = (6 + ) + 3 + 4 49 4 + sin + C 3 7 4 3/ 3 = (6 + ) + (4 ) 6 + 49 4 + sin + C 6 3 7 3/ 3 = (6 + ) + (4 ) 6 637 4 6 + + sin + C 3 7. d 4 + Sol. d 4 + + + = d 4 4 = + + d
+ = + + + 4 4 d + = + d + + + = d + + + tan log = c + + + ( ) ( ) d + = tan log c + + + 3. Find th ara boundd btwn th curvs y 4a, 4by( a 0, b 0) = = > >. Sol: Equations of th givn curvs ar y = 4a.() = 4by..() From quation () y = 4b Substituting in () = 4b 4a ( ) 4 = 6b 4a
3 64b a = 0 X = 0, = 4 ( ) 3 b a Ara boundd will b = ( ) 3 4 b a 0 4a d 4b 3 3 3 4 = ( b a ) ( ) 0 4a. 3 b 3. 3 = ( 4a) 8( b a) 3. ( ) 4 b a 3 b 3 3 = ab 3 6 64.b a b = 3 6 ab 3 3 = 6 ab sq.units 3 4. (y y)d + (y )dy = 0 Sol. (y y)d + (y )dy = 0 (y )dy = (y y)d
dy y y = d y Put dy dv y = v = v + d d v + = d v = (v ) dv v v (v v ) dv v v = v d v v v v + v 3v( v) = = v v v d dv = 3 v( v) () v A B Lt = + v( v) v v v = A( v) + Bv v = 0 = A A = v = = B B = d + dv = 3 v + v log v log( v) = 3log + log c log = log c v( v) 3 = c v( v) = v( v) c 3 y y y y = 3 = 3 c c y( y) = = k y(y ) = = k c c 3