PRACTICE PAPER 6 SOLUTIONS

PRACTICE PAPER 6 SOLUTIONS SECTION A I.. Find the value of k if the points (, ) and (k, 3) are conjugate points with respect to the circle + y 5 + 8y + 6. Sol. Equation of the circle is + y 5 + 8y + 6 Polar of P(, ) is. + y. 5 ( + ) + (y + ) + 6 8 + y 5 + 8y + 6 + 3 + y + 8 P(, ), Q(k, 3) are conjugate points Polar of P passes through Q 3k 36 + 8 8 3k 8 k. 3. If the circle + y + a + by has the centre at (, 3) then find a, b and the radius of the circle. Sol. Equation of the circle is + y + a + by Centre a a, b (, 3) b 3 a b 6 g, f 3, c radius g + f c + 9 + 5 67

3. + y + + 6y 7, ( + y ) + 8 + y 9 find the equation of the radical ais of the circles. Sol. S S' is radical ais. ( + y + + 6y 7) 9 + y + + 3y 9 + 3y 8 + y 9. Find the equations of ais and directri of the parabola y + 6y + 5. Sol. y + 6y 5 Adding '9' on both sides we get, y + 6y + 9 5 + 9 [y ( 3)] + [y ( 3)] [ ( )] Comparing with (y k) a ( h) we get, (h, k) (, 3), a Equation of the ais y k i.e. y + 3 Equation of the directri h + a i.e., ( ) + + 5. 5. If the angle between the asymptotes is 3º then find its eccentricity. Sol. Angle between the asymptotes θ 3º θ 5º b tan θ tan 5º a 68

e a + b a + tan 5º sec 5º 3 + 8 + 3 3 3 8( 3) 8 3 ( 6 ) Eccentricity e 6 6. Solve sec cosec on I R \ { n : n Z} (n + ) : n Z. Mar., May '7 Sol. sec. cosec cos sin sin cos + cos. sin + cos sin sec + cosec tan cot + C cot (log ) 7. Solve, I (, ) \ {e n : n Z). Mar. '5 Sol. t log dt cot (log ) cot t dt log (sin t) + C log (sin (log ) + C 69

8. Evaluate Sol. + ( ) + ( ) + ( 8 8) + 9. Find the area under the curve f() sin in [,]. Sol. Y O / 3/ X f() sin, We know that in [, ], sin and [, ], sin Required area ( cos ) [cos ] sin + ( sin) cos + cos + cos cos ( ) + + ( ) + + +. 7

. Find the general solution of dy y. dy Sol. y dy y log c + log y log log cy log Solution is cy where c is a constant SECTION B II.. If the abscissae of points A, B are the roots of the equation, + a b and ordinates of A, B are roots of y + py q, then find the equation of a circle for which AB is a diameter. Sol. Equation of the circle is ( ) ( ) + (y y ) (y y ) ( + ) + + y y (y + y ) + y y, are roots of + a b y, y are roots of y + py q + a y + y p b y y q a + b + c sum of roots b/a product c/a Equation of circle be ( a) b + y y ( p) q + a + y + py b q 7

. If the angle between the circles + y 6y + and + y + k + 6y 59 is 5 find k. Sol. Suppose θ is the angle between the circles + y 6y + and + y + k + 6y 59 g 6, f 3, c, g k, f 3, c 59 cos θ cos 5 o c + c g. g r r 36 + 9 f. f k 59 ( 6) ( 3). 3 8 + 6k + 8 k.. + 68 6k k. + 68 Squaring and cross - multiplying k + 68 8k [ k 7] + 9k k + 7 8 k 7k 7 k + 9 + 59 7

k 7 k 6 7 k ±. 3. Find the equation of the tangent and normal to the ellipse 9 + 6y at the end of the latus rectum in the first quadrant. Mar. '7 Sol. Given ellipse is 9 + 6y a b e a y + 6 9 6 9 6 7 End of the latus rectum in first quadrant P b 9 ae, a 7, Equation of the tangent at P is yy + a b 7 y 9. +. 6 9 7 y + or 7 + y 6 6 Equation of the normal at P is a b y a b y 6 7 9y 9 6 9 6 y 7 7 6 7y 7 7. 73

. If e, e are the eccentricities of a hyperbola and its conjugate hyperbola prove that +. e e y Sol. Equation of the hyperbola is a b b a (e ) e b a e b a + b + a a a () e a + b Equation of the conjugate hyperbola is y y a b b a a b (e a ) e b a a + b e + b b b () e a + b Adding () and () e a b a + b + + e a + b a + b a + b. 5. Solve + 5sin Mar. '5 Sol. t tan dt sec dt sec. t 7 dt +

sin tan + tan t + t dt dt I t + t + + t t + 5 + t dt 5t + + t dt 5 t + 5 3 t +. log + C 3 5 3. t + + t + t + log + C log + C 3 t + 8 3 t + tan + log + C 3 tan + 6. Evaluate Sol. I / / log (+ tan ) log + tan / tan tan log + + tan tan / tan log + + tan 3 75

/ / + tan + tan log + tan [log log(+ tan ) / / log log (+ tan ) ] / log () I I log I log 8 dy 7. Evaluate tan ( + y). Sol. dy tan ( + y) put v + y dv dy + + tan v sec v dv sec v cos v. dv + c (+ cos v) dv + c ( + cos v) dv + c sin v v + + c v + sin v + c' ( + y) + sin ( + y) + c' y sin [( + y)] c 76

SECTION C III. 8. If (, ), (, ), (, 5) and (, c) are concyclic and then find c. Sol. + y + g + fy + c Satisfies (, ), (, ) (, 5) we get + + g + c (i) + + g. + f + c (ii) 6 + 5 + 8g + f + c (iii) (ii) (i) we get 3 g + f g f 3 (iv) (ii) (iii) we get 8g 8f (or) g + f 5 (v) Solving(iv) and (v) we get 3 7 g, f 6 6 Substituting g and f value in equation (i) we get + 3 + c 6 c 3 Now equation + y 3 7 y + 3 3 3 Now circle passes through (, c) then c 7 c + 3 3 3c 7c + (3c ) (c ) (or) c or. 3 77

9. Show that the locus of the point of intersection of the lines cos α + y sin α a, sin α y cos α b (α is a parameter) is a circle. Sol. Equations of the given lines are cos α + y sin α a sin α y cos α b Let p (, y ) be the point of intersection cos α + y sin α a () sin α y cos α b () Squaring and adding () and () ( cos α + y sin α) + ( sin α y cos α) a + b cos α + y sin α y αsin α + sin α + y + cos cos y cos α sin α a + b (cos α + sin α) + y (sin α + cos ) a + b α y + a + b Locus of p(, y ) is the circle + y a + b. Find the equation of the parabola whose latus rectum is the line segment of joining the points ( 3, ) and ( 3, ). Sol. L ( 3, ) and L' ( 3, ) are the ends of the latus rectum. S is the midpoint of LL' L( 3, ) Co-ordinates of S are 3 3, LL' ( 3 + 3) + ( ) + a, a a ± Case (i) a α S L'( 3, ) 78

Co-ordinates of A are 3 + Equation of the parabola is 3 y + 3 (y 3) ( + ) (y 3) ( + ) Case (ii) a Co-ordinates of A are 79, 3, Equation of the parabola is 3 y + 3 + 3 (y 3) ( + + ) i.e., (y 3) + 3.. Evaluate ( a)( b)( c) Sol. Let ( a)( b)( c) ( a)( b)( c), 3 A a + B b + C c A( b)( c) + B( a)( c) + C( a)( b) ( a)( b)( c) A( b) ( c) + B( a) ( c) + C( a) ( b) () Put a, we get A(a b) (a c) A (a b)(a c)

Put b, we get A() + B(b a) (b c) + C() B (b a)(b c) Similarly C (c a)(c b) ( a)( b)( c) (a b)(a c) (b a)(b c) + + a b ( a)( b)( c) (a b)(a c) (c a)(c b) c + a (b a)(b c) + (c a)(c b) log a + (a b)(a c) +. Evaluate cos + 3sin + 7 cos + sin + Sol. Let cos + 3 sin + 7 c b log b (b a)(b c) log c + k (c a)(c b). A (cos + sin + )' + B(cos + sin + ) + C Comparing the coefficients A + B, A B 3, B + C 7 A, B, C 5 8

cos + 3sin + 7 sin + cos cos + sin + cos + sin + + + 5 cos + sin + log cos + sin + + + 5I...() I cos + sin + cos + sin sec + tan dt + t log + t log + tan cos + 3sin + 7 Substituting in cos + sin + t tan log cos + sin + + + 5 log cos + tan + C 9 3. Solve (9 ) ( ) Sol. Put cos θ + 9 sin θ (9 ) sinθ dθ 5 sinθ dθ U.L. cos θ + 9 sin θ 9 cos θ + 9 sin θ 5 cos θ θ 8

L.L cos θ + 9 sin θ cos θ + 9 sin θ 5 sin θ θ 9 9 ( cos θ + 9 sin θ) (9 ) cos θ 5 cos θ cos θ + 9 sin θ (9 ) sin θ 5 sin θ Let I I 9 (9 ) ( ) 5( sinθ cosθ)dθ 5cos θ5sin θ θ θ dθ θ θ sin cos 5 cos sin. 5 dy 3y 7 + 7. Solve 3 7y 3 Sol. Let + h, y y + k so that dy 3(y + k) 7( + h) + 7 3( + h) 7 (y + k) 3 7h + 3k 3 3h 7k + 7 h k I +3 3 7 3 7 7 3 7 h k 9 + 9 + 9. d () 5 θ 5 dy 9 dy 3y 7 + (3k 7h + 7) (3 7y) + (3h 7k 3) 8

h and k dy 3y 7 3 7y dy dv Put y v v +. dv (3v 7) v +. (3 7v) dv 3v 7 3v 7 3v + 7v 7v 7. v 3 7v 3 7v 3 7v 3 7v 7v 7 3 7v dv dv 7v 7 7v 7 3 v ln ln ln v v + log log c v 3 log 7 log v v + ln ln c 3ln (v ) 3ln (v + ) 7ln (v + ) 7ln (v ) ln ln c ln (v + ) ln (v ) ln (v + ) 5 + ln (v ) + ln 7 ln c (v + ) 5. (v ). 7 c 5 y y +. 7 c (y ) (y + ) 5 c [y ( )] (y + ) 5 c Solution is [y + ] (y + ) 5 c. 7v 7 3 7v 83