Conic Section 2
1) The equation of the parabola is (x+2) = ‐20(y‐3), then the axis is: (1) the x axis (2) the y axis (3) a line parallel to the y axis (4) y=k. Ans: (3) A line parallel to the y axis: It is symmetric about the line x‐2=0 (parallel to y axis). 2) Focus of the parabola y2 + 2y + 4x + 5 = 0 is: (1) (‐2,‐1) (2) (2,‐1) (3) (‐2,1) (4) (2,1) Ans: (1) (‐2,‐1). y2 + 2y + 1 + 4x + 4 = 0 (y+1)2 = ‐4(x+1). Focus is (‐1+h, k) = (‐2, ‐1). 3) The axis of the parabola x2 + 20y + 4x – 56 = 0 is: (1) x+2=0 (2) x‐2=0 (3) y+2=0 (4) y‐2=0. Ans: (1) x+2=0 (x+2)2 = ‐20(y‐3). This axis is x+2 = 0. 4) The line y = 2x+k will intersect the parabola y2 = 2x if: (1) k = ¼ (2) k < ¼. (3) k < ¼ (4) k > ¼ . Ans: (3) k < ¼ . 5) The parametric equation of a parabola is x=t2 +1, y = 2t+1. the Cartesian equation of its directrix is: (1) x=0 (2) x+1 = 0 (3) y=0 (4) y=1. Ans: Eliminating t, x = [(y‐1)/2]2 + 1. Or (y‐1)2 = 4(x‐1). Putting y‐1=Y. x‐1 = X. The equation becomes Y2 = 4x. So, the equation of the directrix is X+1 = 0. x=0. 6) The focus of the parabola y2 – x ‐ 2y + 2 =0 is: (1) (1/4, 0) (2) (1, 2) (3) (3/4, 1) (4) (5/4, 1) Ans: (4) (5/4, 1) (y‐1)2 = 1(x‐1) Focus (a+h, k) = (1/4+1, 1) = (5/4,1). 7) The equation of the tangent to the parabola y2 = 8(x‐1) at (3,4) is: (1) x+y=7 (2) y‐x‐1=0 (3) x‐y+1=0 (4) None. Ans: (3) x‐y+1=0 y.y1 = 4(x+x1 ) ‐ 8 i.e y.4 = 4(x+3) – 8 i.e y = x+3 – 2 i.e x‐y+1 = 0. 8) The point on the parabola y2 = 4xwhich is nearest to the (3,1) is: (1) (1,2) (2) (1,‐2) (3) (4,4) (4) (4,‐4). Ans: (1) (1,2) The required point is the foot of the normal drawn from the point (3,1). If ‘t’ i.e (t2 ,2t) is the required point then, 2t‐1 .1 = ‐1 => t = 1. t2 – 2 t (1,2) is the required point. 9) The equation of the directrix of the parabola y2 + 4y + 4x + 2 = 0 is: (1) x=‐1 (2) x=1 (3) x=‐3/2 (4) x=3/2. Ans: (4) x=3/2 (y+2)2 = ‐4(x ‐ 1/2). Directrix : x = ‐a+h x = 1 +1/2 i.e x =3/2. 10) The equation of the line of symmetry of the parabola 4x2 – 20x – 12y + 49 = 0 is: (1) 2x = 5 (2) x=5. (3) y = 5 (4) None of these Ans: (1) 2x = 5 11) Focus of the parabola y2 – 8x – 32 = 0 is given by: (1) (2,0) (2) (‐2,0) (3) (0,2) (4) (4,0). Ans: Focus of the parabola (y‐k)2 = 4a(x‐h) is given by [a+h,k] here,y2 = 8x + 32. y2 = 8(x+4) so, a = 2, h = ‐4, k=0 Focus is (2‐4, 0) = (‐2,0). 12) Vertex of the parabola y2 – 8x 4y + 4 =0 is at: (1) (2,‐3) (2) (o,2) (3) (‐3,‐2) (4) (‐2,3) Ans: (2) (o,2) Vertex of the parabola (y‐k)2 = 4(x‐h) is (h,k) y2 ‐4y + 4 = 8x (y‐2)2 = 8x s0, h = 0, k = 2 vertex is (0,2). 13) Directrix of the parabola y2 = 16x is : (1) y = 4 (2) x = ‐4 (3) x = 4 (4) y = ‐4. Ans: (2) x = ‐4 directrix of the parabola y2 = 4ax is x = ‐a. Here, y2 = 16x. s0, a = 4 Directrix is ( x = ‐4). 14) Focus of the parabola y2 = 16x: (1) (‐4.0) (2) (4,0) (3) (0,‐4) (4) (0,4). Ans: (2) (4,0) Focus of the parabola y2 = 4ax is (a,0) Here, y2 = 4ax a = 4 Focus (4.0). 15) Equation of the normal of y2 = 20x at (5,10) is: (1) x‐y+5 =0. (2) x+y‐15 =0. (3) x+y‐5 = 0. (4) x‐y+15 = 0. Ans: (2) x+y‐15=0 Eq. of the normal of y2 = 4ax at (x1 , y1 ) is xy1 + 2ay = x1 y1 + 2ay1 . Here, y2 = 20x, a=5: (x1 , y1 ) = (5,10) Then, 10x + 10y = 50 + 100 10x + 10y = 150. x + y = 15. 16) The coordinates of foci of an ellipse (x‐h)2 + (y‐k)2 =1aregivenby a2 b2 (a>b) (1) (h,k) (2) (h+ae,k) (3) (h,k+ae) (4) none of these Ans; (2) (h+ae,k) The coordinates of the foci are given by (h+ae,0+k) (17) Center of the ellipse 9x2+5y2‐36x‐50y‐164 =0 is at (1) (2,5) (2) (1,‐2) (3) (‐2,1) (4) (0,0) Ans; (2,5) The center of ellipse(x‐h)2+(y‐k)2 =1 a2 b2 is at (h,k) Here 9(x2 ‐4x) – 5(y2 ‐10y) = 64 or 9(x‐2)2 + 5(y‐5)2=164+36+12=325 9(x‐2)2+5(y‐5)2 = 1 so center is(2,5). 18)The eccentricity of the ellipse x + y2 =1 16 25 (1) 0.4 (2) 0.5 (3) 0.6 (4) 0.75 Ans; 0.6 Here b2>a2 so a2=b2(1‐e2) so 16=25(1‐e2) or 1‐e2 =16/25 e2=1‐16/25 =9/25 so e=3/5 =0.6. 19)The foci of the ellipse 4x2+3y2=24 are the points (1) (+2,0) (2) (0,+2√2) (3) (0,+√2) (4) (+2√2,0) Ans; (3) (0,+√2) Dividing by 24 equation is x2+ y2=1 6 8 Major axis is the y‐axis, so a2 = b2(1‐e2) So 6 = 8(1‐e2) or 1‐e2 = ¾ => e2 =1/4 or e2=1/2 Foci [ 0,+be]= [0,+√2] 20)The distance between the foci of the ellipse 5x2+9y2=45 is (1) 2 (2) 2√2 (3) 4 (4) 4√2 Ans; The given equation is x2 + y2 =1 9 5 or 5=9(1‐e2) 1‐e2=5/9 or e2=1‐5/9=4/9 so e=2/3 and a=3 so Distance between foci =2ae =2×3×2/3 =4. 21) The four tangents at the end points of latera recta of an ellipse form a: (1) Square (2) parallelogram but not rhombus. (3) Rhombus but not a square. (4) rectangle but not a square. Ans: (4) rectangle but not a square The sides of the rectangle are 2b2 /a and 2ae. 22) the equation of the ellipse with a focus at (0,0) the corresponding directrix: x+6 = 0 and e =1/2 is: (1) 3x2 + 4y2 +12x – 36=0 (2) 3x2 + 4y2 – 12x + 36=0 (3) 3x2 + 4y2 – 12x – 36=0 (4) None of these. Ans: (3) 3x2 + 4y2 – 12x – 36=0 SP = e.PM =>√x2 + y2 = ½[x+6] = 4(x2 + y2 ) = x2 + 12x + 36 Which simplifies to 3x2 + 4y2 – 12x – 36 = 0 23) The length of the LR of the ellipse 49x2 + 64y2 = 3136 is: (1) 49/64 (2) 64/3136. (3) 49/4 (4)none of these. Ans: (3) 49/4 x2 /64 + y2 /49=1Standard form LR = 2b2 /a = 49/4. 24) The equation of the axes of the ellipse 3x2 + 4y2 + 6x – 8y – 50 =0 are : (1) x=0;y=0 (2) x+2=0;y‐2=0 (3) x+1=0;y‐1=0 (4) (x+1)(y+1)=0. Ans: (3) x+1=0;y‐1=0 3(x2 + 2x) + 4(y2 – 2y) = 50. Which is reducible to 3(x+1)2 + 4(y‐1)2 = k2 . The Equation of the axes are x+1=0 ; y‐1=0. 25) The equation of an ellipse is 3x2 + 4y2 = 12 then the length of its latus rectum is: (1) 3/2 (2) 8/3. (3) √3/2 (4) 3 Ans: (4) 3. 2
x /4 +y2 /3 = 1, a2 = 4, b2 = 3, Length of the latus rectum is 2b2 /a = 3. 26) Latus rectum of an ellipse is equal to half of its major axis, its eccentricity is: (1) 1/ √2 (2) ½ (3) 2/3 (4) 4/5. Ans: (1) 1/ √2 27) The equation x2 / (2‐∞) + y2 / (∞‐5) = 1 represents an ellipse if: (1) ∞ > 5 (2) ∞ < 2 (3) 2<∞<5 (4) ∞<2, ∞>5, Ans: (4) ∞<2, ∞>5, 2‐ ∞>0, ∞‐5>0 = 2> ∞ and ∞>5 = ∞<2, ∞>5. 28) Sum of the focal distances of any point on the ellipse 25x2 + 9y2 = 225: (1) 6 (2) 10 (3) 9 (4) 25. Ans: (2) 10. 29) One focus of an ellipse is (3,0) and the corresponding directrix is x‐6=0, its eccentricity is: (1) ‐1/2. (2) ‐2/3. (3) 4/5 (4) 1/√2. Ans: (4) 1/√2. 30) If e = 1/3, then the ratio of the major axis to the minor axis of the ellipse: (1) 3:2√2 (2) 9:8 (3) 2√2:3 (4) 8:9. Ans: (1) 3:2√2 b2 / a2 = 1‐e2 = 1‐ 1/9 = 8/9 b/a = 2√2/3 a/b = 3:2√2 . 31) The equation of a hyperbola is x2 – 4y2 = 4. Its eccentricity is given by: (1) √5 (2) 2 (3) 2/ √5 (4) √5/2. Ans: (4) √5/2. 32) The length of transverse axis of the hyperbola 3x2 – 4y2 = 32 is: (1) 3/32 (2) 64/3 (3) 8 √2/ √3 (4) 16 √2/ √3. Ans: (3) 8 √2/ √3 . 33) In a hyperbola, the distance between the foci is 20 and distance between the directrices is 10. The eccentricity is: (1) √2 (2) 2 (3) 3 (4) √3. Ans: (1) √2 2ae = 20, 2a/e = 10. 10e2 = 20 = √2. 34) The equation of the normal at(2,‐3) on the hyperbola x2–(y2 /3)=1 is: (1) x‐2y=8 (2) 2x‐y=8 (3) x+2y=8 (4) 2x+y=1. Ans: (1) x‐2y=8. 3(y+3))/‐3 = ‐1(x‐2)/2 ‐2y‐6 = ‐x+2 x‐2y‐8 = 0. 35) Equation of the hyperbola whose asymptotes are coordinate axes and passing through the point (8,2) is: (1) x2 – y2 = 60. (2) x.y = 16. 2
2
(3) x + y = 68. (4) x.y = 10. Ans: (2) x.y = 16. Equation of the rectangular hyperbola with coordinates axes as asymptotes is xy=c2 . This passes through (8,2). C2 = 16. Hence the equation is x.y=16. 36) The equation x2 /4‐a ‐ y2 /4+a = 1 represents a hyperbola if: (1) a>4 (2) a>4. (3) ‐4<a<4 (4) a< 4. Ans: (3) ‐4<a<4 4‐a > 0 4 > a. 37) For the hyperbolas x2 / cos2 α – y2 /sin2 α =1 which of the following is always constant? (1) abscissa of the vertices. (2) abscissa of foci. (3) e (4) direct ices. Ans: abscissa of foci. The abscissa of foci = + √a2 + b2 . Becomes = + √ cos2 α+sin2 α. = + 1 so, abscissa of foci. 38) In a hyperbola the length of a latus rectum is equal to the length of the either of the axes. Then its eccentricity is: (1) √2 (2) 2 (3) √2 or 2 (4) √2 or √3/2. Ans: (1) √2 2b2 /a = 2a or 2b2 /a = 2b. b2 / a2 = 1. b/a = 1. a = b so, e = √2. 39) Find the angle between the asymptotes of the hyperbola 3x2 – y2 = 1: 1)1100 2) 1200 4) 1150 3)1400 Ans: 2 tan‐1 b/a = 2 tan‐1 (1/(1/√3) = 2 tan‐1 √3 = 1200 . 40) An ellipse of eccentricity has the same foci as that of the hyperbola y2 /9 ‐ x2 /16 = 1. Then the length of the longest focal chord of the ellipse is: (1) 18/5 (2) 50/9. (3) 25 (4) 10. Ans: (3) 25. Foci of the hyperbola are at (0,+ √a2 + b2 ) i.e (0,+5) e of the ellipse = 0.4 = 2/5 be =5 5 => b.2/5 2b = 25. 41) The eccentricity of the hyperbola 3x2 ‐4y2 =12 is (1) √17 /3 (2) √7 / 2 (3) 4/3 (4) 3/√7 Ans; (2) √7 / 2 Dividing by 12 the hyperbola is x2/4‐y2/3 = 1. b2 = a2(e2‐1) so 3 = 4(e2‐1) or e2‐1 = 3/4 or e2 = 1+3/4 =7/4 or e= √7 / 2. 42) The normal to the hyperbola 4y2 ‐5x2 =20 at the point (‐4,5) is given by (1) y+x=1 (2) y‐x=9 (3) 5y‐4x=41 (4) none of these Ans: (2)y‐x = 9 4y2 ‐5x2 = 20 differentiating 8y dy/dx ‐ 10x = 0 dy/dx=5x/4y or(dy/dx)(‐4,5) = ‐20/20 = ‐1 so slope of Normal = 1 Equation to Normal y‐5 =1(x+4)b or y‐x=9. 43) Eccentricity of the hyperbola 9x2 ‐ 16y2 = 144 is: (1) 5/4 (2) 4/5 (3) 3/5 (4) 4/3. Ans: For the hyperbola x2 / a2 ‐ y2 / b2 = 1: b2 = a2 (e2 – 1). Here, 9x2 – 16y2 = 144. x2 /16 – y2 /9 = 1. so, 9 = 16(e2 – 1) e = 5/4. 44) The centre of the hyperbola 9x2– 16y2 – 18x – 32y =151 is: (1) (1,‐1) (2) (‐1,1) (3) (1,1) (4) (‐1,‐1). Ans: (1) (1,‐1) The equation is 9x2 – 16y2 – 18x – 32y =151 9(x2 – 2x) – 16(y2 – 2y) = 151 9(x‐1)2 – 16(y‐1)2 = 151 +9‐16. 9(x‐1)2 – 16(y‐1)2 = 144. (x‐1)2 / 16 – (y‐1)2 /9 = 11 centre is (1,‐1). 45) The foci of the hyperbola 9y2 –4x2 = 36 are the points: (1) (+3,0) (2) (0,+3) (3) (+√13,0) (4) (0,+√13) Ans: (4) (0,+√13). The given equation is 9y2 – 4x2 = 36 . x2 /9 ‐ y2 /4 = ‐1 a = 3, b=2: b2 = a2 (e2 – 1) 4 = 9 (e2 – 1) e2 = 13/9 e = + √13/3. Foci = (0,+be) = (0, +√13).